1. CHUYN LUYN THI I HC 2013 - 2014 KHO ST HM S BIN SON: LU HUY THNG H NI, 8/2013 H V TN: LP :. TRNG : ketnoitrithuc2013.blogspot.com - Chia s kin thc LTH.

2. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 1 CHUYN : KHO ST S BIN THIN V V TH HM S CC BI TON LIN QUAN N KHO ST HM S VN 1: TNH N IU CA HM S 1. inh ngha: Hm s f ng bin trn 1 2 1 2 1 2( , , ( ) ( ))K x x K x x f x f x < < Hm s f nghch bin trn 1 2 1 2 1 2( , , ( ) ( ))K x x K x x f x f x < > 2. iu kin cn: Gi s f c o hm trn khong I. a) Nu f ng bin trn khong I th '( ) 0,f x x I b) Nu f nghch bin trn khong I th '( ) 0,f x x I 3.iu kin : Gi s f c o hm trn khong I. a) Nu '( ) 0,f x x I ( '( ) 0f x = ti mt s hu hn im) th f ng bin trn I. b) Nu '( ) 0,f x x I ( '( ) 0f x = ti mt s hu hn im) th f nghch bin trn I. c) Nu '( ) 0,f x x I= , x I th f khng i trn I. Ch : Nu khong I c thay bi on hoc na khong th f phi lin tc trn . Dng ton 1: Xt tnh n iu ca hm s Phng php: xt chiu bin thin ca hm s y = f(x), ta thc hin cc bc nh sau: Tm tp xc nh ca hm s. Tnh y. Tm cc im m ti y = 0 hoc y khng tn ti (gi l cc im ti hn) Lp bng xt du y (bng bin thin). T kt lun cc khong ng bin, nghch bin ca hm s. Bi tp c bn HT 1. Xt tnh n iu ca cc hm s sau: 1) 3 2 2 2y x x x= + 2) 2 (4 )( 1)y x x= 3) 3 2 3 4 1y x x x= + 4) 4 21 2 1 4 y x x= 5) 4 2 2 3y x x= + 6) 4 21 1 2 10 10 y x x= + 7) 2 1 5 x y x = + 8) 1 2 x y x = 9) 1 1 1 y x = 10) 3 2 2y x x= + + 11) 2 1 3y x x= 12) 2 2y x x= www.VNMATH.com 3. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 2 Dng ton2: Tm iu kin hm s lun ng bin hoc nghch bin trn tp xc nh (hoc trn tng khong xc nh) Cho hm s ( , )y f x m= , m l tham s, c tp xc nh D. Hm s f ng bin trn D y 0, x D. Hm s f nghch bin trn D y 0, x D. T suy ra iu kin ca m. Ch : 1) y = 0 ch xy ra ti mt s hu hn im. 2) Nu 2 'y ax bx c= + + th: 0 0 ' 0, 0 0 a b c y x R a = = > 0 0 ' 0, 0 0 a b c y x R a = = < 3) nh l v du ca tam thc bc hai 2 ( )g x ax bx c= + + : Nu < 0 th g(x) lun cng du vi a. Nu = 0 th g(x) lun cng du vi a (tr x = 2 b a ) Nu > 0 th g(x) c hai nghim x1, x2 v trong khong hai nghim th g(x) khc du vi a, ngoi khong hai nghim th g(x) cng du vi a. 4) So snh cc nghim 1 2,x x ca tam thc bc hai 2 ( )g x ax bx c= + + vi s 0: 1 2 0 0 0 0 x x P S >< < > < 1 2 0 0 0 0 x x P S >< < > > 1 20 0x x P< < < 5) hm s 3 2 y ax bx cx d= + + + c di khong ng bin (nghch bin) 1 2( ; )x x bng d th ta thc hin cc bc sau: Tnh y. Tm iu kin hm s c khong ng bin v nghch bin: 0 0 a > (1) Bin i 1 2x x d = thnh 2 2 1 2 1 2( ) 4x x x x d+ = (2) S dng nh l Viet a (2) thnh phng trnh theo m. Gii phng trnh, so vi iu kin (1) chn nghim. Bi tp c bn HT 2. Tm m cc hm s sau lun ng bin trn tp xc nh (hoc tng khong xc nh) ca n: 1) 3 2 3 ( 2)y x mx m x m= + + 2) 3 2 2 1 3 2 x mx y x= + 3) x m y x m + = 4) 4mx y x m + = + HT 3. Tm m hm s: 1) 3 2 3y x x mx m= + + + nghch bin trn mt khong c di bng 1. 2) 3 21 1 2 3 1 3 2 y x mx mx m= + + nghch bin trn mt khong c di bng 3. 4. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 3 3) 3 21 ( 1) ( 3) 4 3 y x m x m x= + + + ng bin trn mt khong c di bng 4. HT 4. Tm m hm s: 1) 3 2 ( 1) ( 1) 1 3 x y m x m x= + + + + ng bin trn khong (1; +). 2) 3 2 3(2 1) (12 5) 2y x m x m x= + + + + ng bin trn khong (2; +). 3) 4 ( 2) mx y m x m + = + ng bin trn khong (1; +). 4) x m y x m + = ng bin trong khong (1; +). BI TP TNG HP NNG CAO HT 5. Cho hm s (1).Tm tt c cc gi tr ca tham s m hm s (1) ng bin trn khong . /s: HT 6. Cho hm s c th (Cm).Tm m hm s ng bin trn khong /s: HT 7. Cho hm s . Tm m hm ng bin trn . /s: 5 4 m HT 8. Cho hm s (1), (m l tham s).Tm m hm s (1) ng bin trn khong (1;2). /s: [ ;1)m HT 9. Cho hm s 3 2 3(2 1) (12 5) 2y x m x m x= + + + + ng bin trn khong ( ; 1) v (2; )+ /s: 7 5 12 12 m HT 10. Cho hm s 3 2 2 (2 7 7) 2( 1)(2 3)y x mx m m x m m= + + . Tm m hm s ng bin trn [2; ).+ /s: 5 1 2 m --------------------------------------------------------- 3 2 3 4y x x mx= + ( ;0) 3m x3 2 2 3(2 1) 6 ( 1) 1y m x m m x= + + + + (2; )+ 1m 3 2 (1 2 ) (2 ) 2y x m x m x m= + + + + ( )0;+ 4 2 2 3 1y x mx m= + www.VNMATH.com 5. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 4 VN 2: CC TR CA HM S I. KIN THC CN NH I.Khi nim cc tr ca hm s Gi s hm s f xc nh trn tp ( )D D v 0x D 1) 0x im cc i ca f nu tn ti khong ( ; )a b D v 0 ( ; )x a b sao cho 0( ) ( )f x f x< , { }0( ; )x a b x . Khi 0( )f x c gi l gi tr cc i (cc i) ca f . 2) 0x im cc tiu ca f nu tn ti khong ( ; )a b D v 0 ( ; )x a b sao cho 0( ) ( )f x f x> , { }0( ; )x a b x . Khi 0( )f x c gi l gi tr cc tiu (cc tiu) ca f . 3) Nu 0x l im cc tr ca f th im 0 0( ; ( ))x f x c gi l im cc tr ca th hm s f . II. iu kin cn hm s c cc tr Nu hm s f c o hm ti 0x v t cc tr ti im th 0'( ) 0f x = . Ch : Hm s f ch c th t cc tr ti nhng im m ti o hm bng 0 hoc khng c o hm. III. iu kin hm s c cc tr 1. nh l 1: Gi s hm s f lin tc trn khong ( ; )a b cha im 0x v c o hm trn { }( ; ) oa b x 1) Nu '( )f x i du t m sang dng khi x i qua 0x th f t cc tiu ti 0x . 2) Nu '( )f x i du t dng sang m khi x i qua 0x th f t cc i ti 0x 2. nh l 2: Gi s hm s f c o hm trn khong ( ; )a b cha im 0x , 0'( ) 0f x = v c o hm cp hai khc 0 ti im 0x . 1) Nu 0"( ) 0f x < th f t cc i ti 0x . 2) Nu 0"( ) 0f x > th f t cc tiu ti 0x . II. CC DNG TON Dng ton 1: Tm cc tr ca hm s Qui tc 1: Dng nh l 1. Tm '( )f x . Tm cc im ( 1,2,...)ix i = m ti o hm bng 0 hoc khng c o hm. Xt du '( )f x . Nu '( )f x i du khix i qua ix th hm s t cc tr ti ix . Qui tc 2: Dng nh l 2. Tnh '( )f x Gii phng trnh '( ) 0f x = tm cc nghim ( 1,2,...)ix i = Tnh "( )f x v "( ) ( 1,2,...)if x i = . Nu "( ) 0if x < th hm s t cc i ti ix . Nu "( ) 0if x > th hm s t cc tiu ti ix 6. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 5 Bi tp c bn HT 11. Tm cc tr ca cc hm s sau: 1) 2 3 3 2y x x= 2) 3 2 2 2 1y x x x= + 3) 3 21 4 15 3 y x x x= + 4) 4 2 3 2 x y x= + 5) 4 2 4 5y x x= + 6) 4 2 3 2 2 x y x= + + 7) 2 3 6 2 x x y x + + = + 8) 2 3 4 5 1 x x y x + + = + 9) 2 2 15 3 x x y x = 10) 3 4 ( 2) ( 1)y x x= + 11) 2 2 4 2 1 2 3 x x y x x + = + 12) 2 2 3 4 4 1 x x y x x + + = + + 13) 2 4y x x= 14) 2 2 5y x x= + 15) 2 2y x x x= + Dng ton 2: Tm iu kin hm s c cc tr 1. Nu hm s ( )y f x= t cc tr ti im 0x th 0'( ) 0f x = hoc ti 0x khng c o hm. 2. hm s ( )y f x= ) t cc tr ti im 0x th '( )f x i du khi x i qua 0x . Ch : Hm s bc ba 3 2 y ax bx cx d= + + + c cc tr Phng trnh ' 0y = c hai nghim phn bit. Khi nu x0 l im cc tr th ta c th tnh gi tr cc tr y(x0) bng hai cch: + 3 2 0 0 0 0( )y x ax bx cx d= + + + + 0 0( )y x Ax B= + , trong Ax + B l phn d trong php chia y cho y. Bi tp c bn HT 12. Tm m hm s: 1) 3 2 ( 2) 3 5y m x x mx= + + + c cc i, cc tiu. 2) 3 2 2 3( 1) (2 3 2) ( 1)y x m x m m x m m= + + c cc i, cc tiu. 3) 3 2 2 3 3 3( 1)y x mx m x m= + 4) 3 2 2 3(2 1) 6 ( 1) 1y x m x m m x= + + + + 2x = 5) 3 2 2 3 ( 1) 2y x mx m x= + + t cc i ti 6) 4 2 2( 2) 5y mx m x m= + + c mt cc i 1 . 2 x = www.VNMATH.com 7. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 6 HT 13. Tm , , ,a b c d hm s: 1) 3 2 y ax bx cx d= + + + t cc tiu bng 0 ti 0x = v t cc i bng 4 27 ti 1 3 x = 2) 4 2 y ax bx c= + + c th i qua gc to O v t cc tr bng 9 ti 3x = . HT 14. Tm m cc hm s sau khng c cc tr: 1) 3 2 3 3 3 4y x x mx m= + + + 2) 3 2 3 ( 1) 1y mx mx m x= + HT 15. Tm m hm s : 1) 3 2 2 2 2( 1) ( 4 1) 2( 1)y x m x m m x m= + + + + t cc tr ti hai im 1 2,x x sao cho: 1 2 1 2 1 1 1 ( ) 2 x x x x + = + . 2) 3 21 1 3 y x mx mx= + t cc tr ti hai im 1 2,x x 2 sao cho: 1 2 8x x . 3) 3 21 1 ( 1) 3( 2) 3 3 y mx m x m x= + + t cc tr ti hai im 1 2,x x sao cho: 1 22 1x x+ = . HT 16. Tm m th hm s : 1) 3 2 4y x mx= + c hai im cc tr l A, B v 2 2 900 729 m AB = . 2) 4 2 4y x mx x m= + + c 3 im cc tr l A, B, C v tam gic ABC nhn gc to O lm trng tm. BI TP TNG HP V NNG CAO HT 17. Tm m th hm s : 1) 3 2 2 12 13y x mx x= + c hai im cc tr cch u trc tung. /s: 0m = 2) 3 2 3 3 4y x mx m= + c cc im cc i, cc tiu i xng nhau qua ng phn gic th nht. /s: 1 2 m = 3) 3 2 3 3 4y x mx m= + c cc im cc i, cc tiu v mt pha i vi ng thng : 3 2 8 0d x y + = . /s: { 4 ;10} 3 m HT 18. Tm m th hm s: 1) 3 2 3y x x m= + + c 2 im cc tr ti A, B sao cho 0 120AOB = /s: 12 132 0, 3 m m + = = 2) 4 2 2 2y x mx= + c 3 im cc tr to thnh 1 tam gic c ng trn ngoi tip i qua 3 9 ; 5 5 D /s: 1m = 8. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 7 3) 4 2 2 2y x mx m m= + + + c 3 im cc tr to thnh 1 tam gic c mt gc bng 0 120 . /s: 3 1 3 m = 4) 4 2 4 2 2y x mx m m= + + c 3 im cc tr to thnh 1 tam gic c din tch bng 4. /s: 3 2m = HT 19. Tm m hm s: 1) 3 3 2y x mx= + c hai im cc tr v ng trn qua 2 im cc tr ct ng trn tm (1;1)I bn knh bng 1 ti hai im A, B sao cho din tch tam gic IAB ln nht. /s: 2 3 2 m = 2) 3 2 4 3y x mx x= + c hai im cc tr 1 2,x x tha mn: 1 24 0x x+ = /s: 9 2 m = HT 20. Tm m hm s: 1) 3 2 2 3( 1) 6( 2) 1y x m x m x= + + c ng thng i qua hai im cc tr song song vi ng thng 4 1y x= . /s: 5m = 2) 3 2 2 3( 1) 6 (1 2 )y x m x m m x= + + c cc im cc i, cc tiu ca th nm trn ng thng 4y x= . /s: 1m = 3) 3 2 7 3y x mx x= + + + c ng thng i qua cc im cc i, cc tiu vung gc vi ng thng 3 7y x= . /s: 3 10 2 m = 4) 3 2 2 3y x x m x m= + + c cc im cc i v cc tiu i xng nhau qua ng thng (): 1 5 2 2 y x= . /s: 0m = ------------------------------------------------------- www.VNMATH.com 9. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 8 VN 3: KHO ST S BIN THIN V V TH HM S I. KIN THC CN NH 1. Cc bc kho st s bin thin v v th ca hm s Tm tp xc nh ca hm s. Xt s bin thin ca hm s: + Tm cc gii hn ti v cc, gii hn v cc v tm tim cn (nu c). + Tnh 'y . + Tm cc im ti o hm ' 0y = hoc khng xc nh. + Lp bng bin thin ghi r du ca o hm, chiu bin thin, cc tr ca hm s. V th ca hm s: + Tm im un ca th (i vi hm s bc ba v hm s trng phng). + V cc ng tim cn (nu c) ca th. + Xc nh mt s im c bit ca th nh giao im ca th vi cc trc to (trong trng hp th khng ct cc trc to hoc vic tm to giao im phc tp th c th b qua). C th tm thm mt s im thuc th c th v chnh xc hn. 2. Kho st s bin thin v v th hm bc ba 3 2 ( 0)y ax bx cx d a= + + + Tp xc nh D = . th lun c mt im un v nhn im un lm tm i xng. Cc dng th: a > 0 a < 0 ' 0y = c 2 nghim phn bit 2 ' 3 0b ac = > ' 0y = c nghim kp 2 ' 3 0b ac = = ' 0y = v nghim 2 ' 3 0b ac = < y x0 I y x0 I y x0 I y x0 I 10. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 9 3. Hm s trng phng 4 2 ( 0)y ax bx c a= + + Tp xc nh D = th lun nhn trc tung lm trc i xng. Cc dng th: 4. Hm s nht bin ( 0; 0) ax b y c ad bc cx d + = + Tp xc nh D =d c th c mt tim cn ng l v mt tim cn ngang l . Giao im ca hai tim cn l tm i xng ca th hm s. Cc dng th: Bi tp c bn HT 21. Kho st s bin thin v v th cc hm s sau: 1. 3 2 3 1y x x= + 2. 3 2 1 3 x y x x= + 3. 3 2 2 1 3 x y x x= + + 4. 4 2 2 2y x x= + 5. 4 2 1y x x= + 6. 1 1 x y x = + 7. 2 1 1 x y x = 8. 1 2 1 x y x = + ---------------------------------------------------- d x c = a y c = a > 0 a < 0 c 3 nghim phn bit ch c 1 nghim y x0 y x0 y x0 y x0 0 ad bc > x y 0 ad bc < x y www.VNMATH.com 11. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 10 VN 4: S TNG GIAO CA CC TH Dng ton 1: Dng th hm s bin lun s nghim phng trnh C s ca phng php: Xt phng trnh: ( ) ( )f x g x= (1) S nghim ca phng trnh (1) = S giao im ca 1( ) : ( )C y f x= v 2( ) : ( )C y g x= Nghim ca phng trnh (1) l honh giao im ca 1( ) : ( )C y f x= v 2( ) : ( )C y g x= bin lun s nghim ca phng trnh ( , ) 0F x m = (*) bng th ta bin i (*) v dng sau: ( , ) 0 ( ) ( )F x m f x g m= = (1) Khi (1) c th xem l phng trnh honh giao im ca hai ng: ( ) : ( )C y f x= v : ( )d y g m= d l ng thng cng phng vi trc honh. Da vo th (C) ta bin lun s giao im ca (C) v d . T suy ra s nghim ca (1) Bi tp c bn HT 22. Kho st s bin thin v v th (C) ca hm s. Dng th (C) bin lun theo m s nghim ca phng trnh: 1) 3 3 3 1; 3 1 0y x x x x m= + + = 2) 3 3 3 1; 3 1 0y x x x x m= + + + = 3) 3 3 2 3 1; 3 2 2 0y x x x x m m= + = 4) 3 3 3 1; 3 4 0y x x x x m= + + + = 5) 4 2 4 2 2 2; 4 4 2 0 2 x y x x x m= + + + = 6) 4 2 4 2 2 2; 2 2 0y x x x x m= + + = HT 23. Kho st s bin thin v v th (C) ca hm s. Dng th (C) bin lun theo m s nghim ca phng trnh: 1) 3 2 3 2 ( ) : 3 6; 3 6 3 0C y x x x x m= + + + = 2) 33 2 2 ( ) : 2 9 12 4; 2 9 12 0C y x x x x x x m= + + + = 3) 2 2 2 ( ) : ( 1) (2 ); ( 1) 2 ( 1) (2 )C y x x x x m m= + + = + 4) 1 11 1 1 ( ) : ; ; ; 1 1 1 1 1 x xx x x C y m m m m x x x x x = = = = = + + + + + ------------------------------------------------ y x g(m A (C) (4) : y = g(m)yC yCT xA 12. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 11 Dng ton 2: Tm iu kin tng giao gia cc th 1.Cho hai th 1( ) : ( )C y f x= v 2( ) : ( )C y g x= . tm honh giao im ca 1( )C v 2( )C ta gii phng trnh: ( ) ( )f x g x= (*) (gi l phng trnh honh giao im). S nghim ca phng trnh (*) bng s giao im ca hai th. 2. th hm s bc ba 3 2 ( 0)y ax bx cx d a= + + + ct trc honh ti 3 im phn bit Phng trnh 3 2 0ax bx cx d+ + + = c 3 nghim phn bit. Bi tp c bn HT 24. Tm to giao im ca cc th ca cc hm s sau: 1) 2 3 3 2 2 1 2 2 x y x x y = + = + 2) 2 2 4 1 2 4 x y x y x x = = + + 3) 3 4 3 2 y x x y x = = + HT 25. Tm m th cc hm s: 1) 2 2 ( 1)( 3)y x x mx m= + ct trc honh ti ba im phn bit. 2) 3 2 3 (1 2 ) 1y mx mx m x= + ct trc honh ti ba im phn bit. 3) 3 2 2 2 ; 2y x x mx m y x= + + + = + ct nhau ti ba im phn bit. 4) 3 2 2 2 2 2 1; 2 2y x x x m y x x= + + = + ct nhau ti ba im phn bit. HT 26. Tm m th cc hm s: 1) 4 2 2 1;y x x y m= = ct nhau ti bn im phn bit. 2) 4 2 3 ( 1)y x m m x m= + + ct trc honh ti bn im phn bit. 3) 4 2 2 (2 3) 3y x m x m m= + ct trc honh ti bn im phn bit. HT 27. Bin lun theo m s giao im ca cc th ca cc hm s sau: 1) 3 3 2 ( 2) y x x y m x = = 2) 3 3 3 ( 3) x y x y m x = + = HT 28. Tm m th ca cc hm s: 1) 3 1 ; 2 4 x y y x m x + = = + ct nhau ti hai im phn bit A, B. Khi tm m on AB ngn nht. 2) 4 1 ; 2 x y y x m x = = + ct nhau ti hai im phn bit A, B. Khi tm m on AB ngn nht. www.VNMATH.com 13. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 12 BI TP TNG HP V NNG CAO HT 29. Tm m hm s: 1) 2 1 ( ) 1 x y C x = + ct ng thng : y x m = + ti hai im phn bit A, B sao cho 2 2AB = /s: 1; 7m m= = 2) 1 ( ) 2 x y C x = ct ng thng : y x m = + ti hai im phn bit A, B sao cho A, B c di nh nht. /s: 1 2 m = 3) 2 1 ( ) 1 x y C x = ct ng thng : y x m = + ti hai im phn bit A, B sao cho tam gic OAB vung ti O. /s: 2m = 4) 2 2 3 ( ) 2 mx m y C x = + ct ng thng : 2y x = ti hai im phn bit A, B sao cho 0 45AOB = 5) (1 ) 2(1 )m x m y x + + = ct ng thng : y x = ti hai im phn bit A, B sao cho: 4 OA OB OB OA + = 6) 3 1 1 x y x + = ct ng thng : ( 1) 2y m x m = + + ti hai im phn bit A, B sao cho tam gic OAB c din tch bng 3 . 2 7) 1 ( ) 2 1 x y C x + = + ct ng thng : 2 2 1 0,mx y m + + = ct th (C) ti hai im phn bit A, B sao cho biu thc 2 2 P OA OB= + t gi tr nh nht. HT 30. Cho hm s 2 ( ) 1 x y C x + = Gi I l giao im ca hai tim cn. Tm trn th (C) hai im A, B sao cho tam gic IAB nhn (4; 2)H lm trc tm. /s: (2;4),( 2;0) HT 31. Cho hm s 1 ( ) 2 1 x y C x + = Xc nh m ng thng : y x m = + ct th (C) ti hai im phn bit c honh 1 2,x x sao cho tng 1 2'( ) '( )f x f x+ t gi tr ln nht. HT 32. Cho hm s 1 ( ) 2 1 x y C x = + Xc nh m ng thng : y x m = + ct th (C) ti hai im phn bit c honh 1 2,x x sao cho tng 1 2'( ) '( )f x f x+ t gi tr nh nht. HT 33. Cho hm s 3 4 ( ) 2 3 x y C x = Xc nh ta cc im trn th (C) sao cho tng khong cch t im n trc honh gp 2 ln khong cch t im n tim cn ng. ----------------------------------------------------- 14. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 13 VN 5: S TIP XC CA HAI NG CONG 1. ngha hnh hc ca o hm: o hm ca hm s ( )y f x= ti im 0x l h s gc ca tip tuyn vi th (C) ca hm s ti im ( )0 0 0; ( )M x f x . Khi phng trnh tip tuyn ca (C) ti im ( )0 0 0; ( )M x f x l: 0 0 0'( )( )y y f x x x = 0 0( ( ))y f x= 2. iu kin cn v hai ng 1( ) : ( )C y f x= v 2( ) : ( )C y g x= tip xc nhau l h phng trnh sau c nghim: ( ) ( ) '( ) '( ) f x g x f x g x = = (*) Nghim ca h (*) l honh ca tip im ca hai ng . 3. Nu 1( ) :C y px q= + v 2 2( ) :C y ax bx c= + + th (C1) v (C2) tip xc nhau phng trnh 2 ax bx c px q+ + = + c nghim kp. Dng ton 1: Lp phng trnh tip tuyn ca ng cong (C): y = f(x) Bi ton 1: Vit phng trnh tip tuyn ca( ) : ( )C y f x= ti im ( )0 0 0;M x y : Nu cho 0x th tm 0 0( )y f x= Nu cho 0y th tm 0x l nghim ca phng trnh 0( )f x y= . Tnh ' '( )y f x= . Suy ra 0 0'( ) '( )y x f x= . Phng trnh tip tuyn l: 0 0 0'( )( )y y f x x x = Bi ton 2: Vit phng trnh tip tuyn ca ( ) : ( )C y f x= bit c h s gc k cho trc. Cch 1: Tm to tip im. Gi M(x0; y0) l tip im. Tnh f (x0). c h s gc k f (x0) = k (1) Gii phng trnh (1), tm c x0 v tnh y0 = f(x0). T vit phng trnh ca . Cch 2: Dng iu kin tip xc. Phng trnh ng thng c dng: y = kx + m. tip xc vi (C) khi v ch khi h phng trnh sau c nghim: ( ) '( ) f x kx m f x k = + = (*) www.VNMATH.com 15. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 14 Gii h (*), tm c m. T vit phng trnh ca . Ch : H s gc k ca tip tuyn c th c cho gin tip nh sau: + to vi chiu dng trc honh gc th k = tan + song song vi ng thng d: y = ax + b th k = a + vung gc vi ng thng d: y = ax + b (a 0) th k = 1 a + to vi ng thng d: y = ax + b mt gc th tan 1 k a ka = + Bi ton 3: Vit phng trnh tip tuyn ca (C): y = f(x), bit i qua im ( ; )A AA x y . Cch 1:Tm to tip im. Gi M(x0; y0) l tip im. Khi : y0 = f(x0), y0 = f (x0). Phng trnh tip tuyn ti M: y y0 = f (x0).(x x0) i qua ( ; )A AA x y nn: yA y0 = f (x0).(xA x0) (2) Gii phng trnh (2), tm c x0. T vit phng trnh ca . Cch 2: Dng iu kin tip xc. Phng trnh ng thng i qua ( ; )A AA x y v c h s gc k: y yA = k(x x1) tip xc vi (C) khi v ch khi h phng trnh sau c nghim: ( ) ( ) '( ) A Af x k x x y f x k = + = (*) Gii h (*), tm c x (suy ra k). T vit phng trnh tip tuyn . Bi tp c bn HT 34. Vit phng trnh tip tuyn ca (C) ti im c ch ra: 1) 3 2 ( ) : 3 7 1C y x x x= + ti A(0; 1) 2) ( ) :C 4 2 2 1y x x= + ti B(1; 0) 3) (C): 3 4 2 3 x y x + = ti C(1; 7) 4) (C): 1 2 x y x + = ti cc giao im ca (C) vi trc honh, trc tung. 5) (C): 2 2 2 1y x x= + ti cc giao im ca (C) vi trc honh, trc tung. 6) (C): 3 3 1y x x= + ti im un ca (C). HT 35. Vit phng trnh tip tuyn ca (C) ti cc giao im ca (C) vi ng c ch ra: 1) (C): 3 2 2 3 9 4y x x x= + v d: 7 4y x= + . 16. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 15 2) (C): 3 2 2 3 9 4y x x x= + v (P): 2 8 3y x x= + . HT 36. Tnh din tch tam gic chn hai trc to bi tip tuyn ca th (C) ti im c ch ra: (C): 5 11 2 3 x y x + = ti im A c xA = 2 . HT 37. Tm m tip tuyn ca th (C) ti im c ch ra chn hai trc to mt tam gic c din tch bng S cho trc: 1) (C): 2 1 x m y x + = ti im A c xA = 2 v 1 2 S = . 2) (C): 3 2 x m y x = + ti im B c xB = 1 v S = 9 2 . 3) (C): 3 1 ( 1)y x m x= + + ti im C c xC = 0 v S = 8. HT 38. Vit phng trnh tip tuyn ca (C), bit c h s gc k c ch ra: 1) (C): 3 2 2 3 5y x x= + ; k = 12 2) (C): 2 1 2 x y x = ; k = 3 HT 39. Vit phng trnh tip tuyn ca (C), bit song song vi ng thng d cho trc: 1) (C): 3 2 2 3 1 3 x y x x= + + ; d: y = 3x + 2 2) (C): 2 1 2 x y x = ; d: 3 2 4 y x= + HT 40. Vit phng trnh tip tuyn ca (C), bit vung gc vi ng thng d cho trc: 1) (C): 3 2 2 3 1 3 x y x x= + + ; d: 2 8 x y = + 2) (C): 2 1 2 x y x = ; d: y x= HT 41. Tm m tip tuyn ca (C) ti im c ch ra song song vi ng thng d cho trc: 1) (C): 2 (3 1) ( 0) m x m m y m x m + + = + ti im A c yA = 0 v d: 10y x= . HT 42. Vit phng trnh tip tuyn ca (C), bit i qua im c ch ra: 1) (C): 3 3 2y x x= + ; A(2; 4) 2) (C): 3 3 1y x x= + ; B(1; 6) 3) (C): ( ) 2 2 2y x= ; C(0; 4) 4) (C): 4 21 3 3 2 2 y x x= + ; 3 0; 2 D 5) (C): 2 2 x y x + = ; E(6; 5) 6) (C): 3 4 1 x y x + = ; F(2; 3) HT 43. Tm m hai ng (C1), (C2) tip xc nhau: 1) 3 2 1 2( ) : (3 ) 2; ( ) :C y x m x mx C= + + + trc honh 2) 3 2 1 2( ) : 2 ( 1) ; ( ) :C y x x m x m C= + trc honh 3) 3 1 2( ) : ( 1) 1; ( ) : 1C y x m x C y x= + + + = + 4) 3 2 1 2( ) : 2 2 1; ( ) :C y x x x C y x m= + + = + HT 44. Tm m hai ng (C1), (C2) tip xc nhau: 1) 4 2 2 1 2( ) : 2 1; ( ) : 2C y x x C y mx m= + + = + www.VNMATH.com 17. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 16 2) 4 2 2 1 2( ) : 1; ( ) :C y x x C y x m= + = + 3) 4 2 2 1 2 1 9 ( ) : 2 ; ( ) : 4 4 C y x x C y x m= + + = + 4) 2 2 2 1 2( ) : ( 1) ( 1) ; ( ) : 2C y x x C y x m= + = + 5) 2 1 2 (2 1) ( ) : ; ( ) : 1 m x m C y C y x x = = Dng ton 2: Tm nhng im trn ng thng d m t c th v c 1, 2, 3, tip tuyn vi th (C): ( )y f x= Gi s d: ax + by +c = 0. M(xM; yM) d. Phng trnh ng thng qua M c h s gc k: y = k(x xM) + yM tip xc vi (C) khi h sau c nghim: ( ) ( ) (1) '( ) (2) M Mf x k x x y f x k = + = Th k t (2) vo (1) ta c: f(x) = (x xM).f (x) + yM (C) S tip tuyn ca (C) v t M = S nghim x ca (C) Bi tp c bn HT 45. Tm cc im trn th (C) m t v c ng mt tip tuyn vi (C): 1) 3 2 ( ) : 3 2C y x x= + 2) 3 ( ) : 3 1C y x x= + HT 46. Tm cc im trn ng thng d m t v c ng mt tip tuyn vi (C): 1) 1 ( ) : 1 x C y x + = ; d l trc tung 2) 3 ( ) : 1 x C y x + = ; d: y = 2x + 1 HT 47. Tm cc im trn ng thng d m t v c t nht mt tip tuyn vi (C): 1) 2 1 ( ) : 2 x C y x + = ; d: x = 3 2) 3 4 ( ) : 4 3 x C y x + = ; d: y = 2 HT 48. Tm cc im trn ng thng d m t v c ba tip tuyn vi (C): 1) 3 2 ( ) : 3 2C y x x= + ; d: y = 2 2) 3 ( ) : 3C y x x= ; d: x = 2 3) 3 ( ) : 3 2C y x x= + + ; d l trc honh 4) 3 ( ) : 12 12C y x x= + ; d: y = 4 HT 49. T im A c th k c bao nhiu tip tuyn vi (C): 1) 3 2 ( ) : 9 17 2C y x x x= + + ; A(2; 5) 2) 3 21 4 4 ( ) : 2 3 4; ; 3 9 3 C y x x x A = + + HT 50. T mt im bt k trn ng thng d c th k c bao nhiu tip tuyn vi (C): 1) 3 2 ( ) : 6 9 1C y x x x= + ; : 2d x = 2) 3 ( ) : 3C y x x= ; : 2d x = 18. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 17 Dng ton 3: Tm nhng im m t c th v c 2 tip tuyn vi th (C): y = f(x) v 2 tip tuyn vung gc vi nhau Gi M(xM; yM). Phng trnh ng thng qua M c h s gc k: y = k(x xM) + yM tip xc vi (C) khi h sau c nghim: ( ) ( ) (1) '( ) (2) M Mf x k x x y f x k = + = Th k t (2) vo (1) ta c: f(x) = (x xM).f (x) + yM (C) Qua M v c 2 tip tuyn vi (C) (C) c 2 nghim phn bit x1, x2. Hai tip tuyn vung gc vi nhau f (x1).f (x2) = 1 T tm c M. Ch : Qua M v c 2 tip tuyn vi (C) sao cho 2 tip im nm v hai pha vi trc honh th 1 2 (3) 2 ( ). ( ) 0 co nghiem phan biet f x f x < Bi tp c bn HT 51. Chng minh rng t im A lun k c hai tip tuyn vi (C) vung gc vi nhau. Vit phng trnh cc tip tuyn : 2 1 ( ) : 2 3 1; 0; 4 C y x x A = + HT 52. Tm cc im trn ng thng d m t c th v c hai tip tuyn vi (C) vung gc vi nhau: 1) 3 2 ( ) : 3 2C y x x= + ; d: y = 2 2) 3 2 ( ) : 3C y x x= + ; d l trc honh Dng ton 4: Cc bi ton khc v tip tuyn HT 53. Cho hypebol (H) v im M bt k thuc (H). Gi I l giao im ca hai tim cn. Tip tuyn ti M ct 2 tim cn ti A v B. 1) Chng minh M l trung im ca on AB. 2) Chng minh din tch ca IAB l mt hng s. 3) Tm im M chu vi IAB l nh nht. 4) Tm M bn knh, chu vi, din tch ng trn ngoi tip tam gic IAB t gi tr nh nht. 5) Tm M bn knh, chu vi, din tch ng trn ni tip tam gic IAB t gi tr ln nht. 6) Tm M khong cch t I n tip tuyn l ln nht. 1) 2 1 ( ) : 1 x H y x = 2) 1 ( ) : 1 x H y x + = 3) 4 5 ( ) : 2 3 x H y x = + HT 54. Tm m tip tuyn ti im M bt k thuc hypebol (H) ct hai ng tim cn to thnh mt tam gic c din tch bng S: 1) 2 3 ( ) : ; 8 mx H y S x m + = = www.VNMATH.com 19. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 18 VN 7: KHONG CCH Kin thc c bn: 1) Khong cch gia hai im A, B: AB = 2 2 ( ) ( )B A B Ax x y y + 2) Khong cch t im M(x0; y0) n ng thng : ax + by + c = 0: d(M, ) = 0 0 2 2 ax by c a b + + + 3) Din tch tam gic ABC: S = ( ) 2 2 21 1 . .sin . . 2 2 AB AC A AB AC AB AC= Bi tp c bn HT 55. Tm cc im M thuc hypebol (H) sao cho tng cc khong cch t n hai tim cn l nh nht. 1) 2 ( ) : 2 x H y x + = 2) 2 1 ( ) : 1 x H y x = + 3) 4 9 ( ) : 3 x H y x = HT 56. Tm cc im M thuc hypebol (H) sao cho tng cc khong cch t n hai trc to l nh nht. 1) 1 ( ) : 1 x H y x = + 2) 2 1 ( ) : 2 x H y x + = 3) 4 9 ( ) : 3 x H y x = HT 57. Cho hypebol (H). Tm hai im A, B thuc hai nhnh khc nhau ca (H) sao cho di AB l nh nht. 1) 1 ( ) : 1 x H y x = + 2) 2 3 ( ) : 2 x H y x + = 3) 4 9 ( ) : 3 x H y x = HT 58. Cho (C) v ng thng d. Tm m d ct (C) ti 2 im A, B sao cho di AB l nh nht. 1 ( ) : ; : 2 0 1 x H y d x y m x + = + = 20. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 19 N TP TNG HP PHN I: TNH N IU CA HM S HT 1. Cho hm s 3 21 ( 1) (3 2) 3 y m x mx m x= + + (1).Tm tt c cc gi tr ca tham s m hm s (1) ng bin trn tp xc nh ca n. /s: 2m HT 2. Cho hm s 3 2 3 4y x x mx= + (1).Tm tt c cc gi tr ca tham s m hm s (1) ng bin trn khong ( ; 0) . /s: 3m HT 3. Cho hm s x3 2 2 3(2 1) 6 ( 1) 1y m x m m x= + + + + c th (Cm).Tm m hm s ng bin trn khong (2; )+ /s: 1m HT 4. Cho hm s 3 2 (1 2 ) (2 ) 2y x m x m x m= + + + + . Tm m hm ng bin trn ( )0;+ . /s: 5 4 m HT 5. Cho hm s 4 2 2 3 1y x mx m= + (1), (m l tham s).Tm m hm s (1) ng bin trn khong (1; 2). /s: ( ;1m . HT 6. Cho hm s 4mx y x m + = + (1). Tm tt c cc gi tr ca tham s m hm s (1) nghch bin trn khong ( ;1) ./s: 2 1m < . HT 7. Cho hm s 3 2 3y x x mx m= + + + . Tm m hm s nghch bin trn on c di bng 1. /s: 9 4 m = PHN II: CC TR CA HM S HT 8. Cho hm s 3 2 (1 2 ) (2 ) 2y x m x m x m= + + + + (m l tham s) (1). Tm cc gi tr ca m th hm s (1) c im cc i, im cc tiu, ng thi honh ca im cc tiu nh hn 1. /s: 5 7 4 5 m< < . HT 9. Cho hm s 3 2 ( 2) 3 5y m x x mx= + + + , m l tham s.Tm cc gi tr ca m cc im cc i, cc tiu ca th hm s cho c honh l cc s dng. /s: 3 2m < < HT 10. Cho hm s 3 2 3 2 3( 2) 6(5 1) (4 2).y x m x m x m= + + + + Tm m hm s t cc tiu ti (0 1;2x /s: 1 0 3 m < HT 11. Cho hm s 4 21 3 2 2 y x mx= + (1).Xc nh m th ca hm s (1) c cc tiu m khng c cc i. /s: 0m HT 12. Cho hm s 4 2 2 4 ( ).my x mx C= + Tm cc gi tr ca m tt c cc im cc tr ca ( )mC u nm trn cc trc ta . /s: 2; 0m m= HT 13. Cho hm s 3 2 2 (2 1) ( 3 2) 4y x m x m m x= + + + (m l tham s) c th l (Cm). Xc nh m (Cm) c cc im cc i v cc tiu nm v hai pha ca trc tung. /s:1 2m< < . HT 14. Cho hm s 3 21 (2 1) 3 3 y x mx m x= + (m l tham s) c th l (Cm). Xc nh m (Cm) c cc im cc i, cc tiu nm v cng mt pha i vi trc tung. /s: 1 1 2 m m > HT 15. Cho hm s 3 2 3 2y x x mx m= + + + (m l tham s) c th l (Cm). Xc nh m (Cm) c cc im cc i www.VNMATH.com 21. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 20 v cc tiu nm v hai pha i vi trc honh. /s: 3m < HT 16. Cho hm s 3 2 31 4 ( 1) ( 1) ( ). 3 3 y x m x m C= + + + Tm m cc im cc tr ca hm s (C) nm v hai pha (pha trong v pha ngoi) ca ng trn c phng trnh: 2 2 4 3 0.x y x+ + = /s: 1 2 m < HT 17. Cho hm s 3 2 3 3 4y x mx m= + (m l tham s) c th l (Cm). Xc nh m (Cm) c cc im cc i v cc tiu i xng nhau qua ng thng y = x. /s: 2 2 m = HT 18. Cho hm s 3 2 3 3 1y x mx m= + . Vi gi tr no ca m th th hm s c im cc i v im cc tiu i xng vi nhau qua ng thng : 8 74 0d x y+ = . /s: 2m = HT 19. Cho hm s 3 2 2 3 2 3 3(1 )y x mx m x m m= + + + (1).Vit phng trnh ng thng qua hai im cc tr ca th hm s (1). /s: 2 2y x m m= + . HT 20. Cho hm s 3 2 3 2 ( ).my x x mx C= + + Tm m ( )mC c cc i v cc tiu, ng thi cc im cc tr ca hm s cch u ng thng : 1 0.d x y = /s: 0m = HT 21. Cho hm s 3 2 3 2y x x mx= + (m l tham s) c th l (Cm). Xc nh m (Cm) c cc im cc i v cc tiu cch u ng thng 1y x= . /s: 3 0; 2 m = HT 22. Cho hm s 3 2 3y x x mx= + (1). Vi gi tr no ca m th th hm s (1) c cc im cc i v im cc tiu i xng vi nhau qua ng thng : 2 5 0d x y = . /s: 0m = HT 23. Cho hm s 3 2 3( 1) 9 2y x m x x m= + + + (1) c th l (Cm). Vi gi tr no ca m th th hm s c im cc i v im cc tiu i xng vi nhau qua ng thng 1 : 2 d y x= . /s: 1m = . HT 24. Cho hm s 3 21 1 ( 1) 3( 2) 3 3 y x m x m x= + + , vi m l tham s thc. Xc nh m hm s cho t cc tr ti 1 2,x x sao cho 1 22 1x x+ = . /s: 4 34 4 m = . HT 25. Cho hm s 3 2 3( 1) 9y x m x x m= + + , vi m l tham s thc. Xc nh m hm s cho t cc tr ti 1 2,x x sao cho 1 2 2x x ./s: 3 1 3m < v 1 3 1.m + < HT 26. Cho hm s 3 2 (1 2 ) (2 ) 2y x m x m x m= + + + + , vi m l tham s thc. Xc nh m hm s cho t cc tr ti 1 2,x x sao cho 1 2 1 3 x x > ./s: 3 29 1 8 m m + > < HT 27. Cho hm s 3 2 4 3y x mx x= + . Tm m hm s c hai im cc tr 1 2,x x tha 1 24x x= . /s: 9 2 m = HT 28. Tm cc gi tr ca m hm s 3 2 21 1 ( 3) 3 2 y x mx m x= + c cc i 1x , cc tiu 2x ng thi 1x ; 2x l di cc cnh gc vung ca mt tam gic vung c di cnh huyn bng 5 2 /s: 14 2 m = HT 29. Cho hm s 3 2 22 ( 1) ( 4 3) 1. 3 y x m x m m x= + + + + + + Tm m hm s c cc tr. Tm gi tr ln nht ca biu thc 1 2 1 22( )A x x x x= + vi 1 2,x x l cc im cc tr ca hm s. 22. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 21 /s: 9 2 A khi 4m = HT 30. Cho hm s 3 2 3( 1) 9 (1)y x m x x m= + + vi m l tham s thc. Xc nh m hm s (1) t cc i , cc tiu sao cho 2CD CTy y+ = /s: 1 3 m m = = HT 31. Cho hm s (C3 2 21 ( 1) 1 ). 3 my x mx m x= + + Tm m hm s c cc i cc tiu v: D 2C CTy y+ > /s: 1 0 1 m m < < > HT 32. Cho hm s 3 2 3 2y x x= + (1). Tm im M thuc ng thng : 3 2d y x= sao tng khong cch t M ti hai im cc tr nh nht. /s: 4 2 ; 5 5 M HT 33. Cho hm s 3 2 2 3 3 3( 1)y x mx m x m m= + + (1). Tm m hm s (1) c cc tr ng thi khong cch t im cc i ca th hm s n gc ta O bng 2 ln khong cch t im cc tiu ca th hm s n gc ta O. /s: 3 2 2 3 2 2 m m = + = . HT 34. Cho hm s 3 2 3( 1) 3 ( 2) 2 ( )y x m x m m x m C= + + + + .Tm m th hm s (C) c cc tr ng thi khong cch t im cc i ca th hm s (C) ti trc Ox bng khong cch t im cc tiu ca th hm s (C) ti trc .Oy /s: 2; 1; 1; 0m m m m= = = = HT 35. Cho hm s 3 2 3 2y x x mx= + c th l (Cm). Tm m (Cm) c cc im cc i, cc tiu v ng thng i qua cc im cc tr song song vi ng thng : 4 3d y x= + ./s: 3m = HT 36. Cho hm s 3 2 3 2y x x mx= + c th l (Cm). Tm m (Cm) c cc im cc i, cc tiu v ng thng i qua cc im cc tr to vi ng thng : 4 5 0d x y+ = mt gc 0 45 . /s: 1 2 m = HT 37. Cho hm s 3 2 3y x x m= + + (1).Xc nh m th ca hm s (1) c hai im cc tr A, B sao cho 0 120AOB = . /s: 12 2 3 3 m + = HT 38. Cho hm s 3 2 2 3 3 3( 1) 4 1 (1),y x mx m x m m m= + + l tham s thc. Tm cc gi tr ca m th hm s (1) c hai im cc tr ,A B sao cho tam gic OAB vung ti ,O vi O l gc ta . /s: 1; 2m m= = HT 39. Cho hm s 3 2 3 2 3( 1) 3 ( 2) 3 .y x m x m m x m m= + + + + + + Chng minh rng vi mi m hm s lun c 2 cc tr v khong cch gia hai im ny khng ph thuc vo v tr ca m. HT 40. Cho hm s 3 2 3 2y x x mx= + (1) vi m l tham s thc. nh m hm s (1) c cc tr, ng thi ng thng i qua hai im cc tr ca th hm s to vi hai trc ta mt tam gic cn. /s: 3 2 m = HT 41. Cho hm s 4 2 2 ( ) 2( 2) 5 5y f x x m x m m= = + + + ( )mC . Tm cc gi tr ca m th ( )mC ca hm s c cc im cc i, cc tiu to thnh 1 tam gic vung cn. /s: 1m = HT 42. Cho hm s ( )4 2 2 2( 2) 5 5 .my x m x m m C= + + + Vi nhng gi tr no ca m th th (Cm) c im cc i v im cc tiu, ng thi cc im cc i v im cc tiu lp thnh mt tam gic u. /s: www.VNMATH.com 23. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 22 3 2 3m = . HT 43. Cho hm s 4 2 2 2y x mx m m= + + + c th (Cm) . Vi nhng gi tr no ca m th th (Cm) c ba im cc tr, ng thi ba im cc tr lp thnh mt tam gic c mt gc bng 0 120 . /s: 3 1 3 m = . HT 44. Cho hm s 4 2 2 1y x mx m= + c th (Cm) . Vi nhng gi tr no ca m th th (Cm) c ba im cc tr, ng thi ba im cc tr lp thnh mt tam gic c bn knh ng trn ngoi tip bng 1 . /s: 5 1 1; 2 m m = = HT 45. Cho hm s 4 2 4 2 2y x mx m m= + + c th (Cm) . Vi nhng gi tr no ca m th th (Cm) c ba im cc tr, ng thi ba im cc tr lp thnh mt tam gic c din tch bng 4. /s: 5 16m = . HT 46. Cho hm s 4 2 2 2x mx + c th ( )mC . Tm tt c cc gi tr ca tham s m th ( )mC c ba im cc tr to thnh mt tam gic c ng trn ngoi tip i qua im D 3 9 ; 5 5 /s: m = 1 PHN 3: S TNG GIAO HT 47. Cho hm s 3 2 6 9 6y x x x= + c th l (C). nh m ng thng ( ) : 2 4d y mx m= ct th (C) ti ba im phn bit. /s: 3m > HT 48. Cho hm s 3 2 3 2y x m x m= (Cm). Tm m (Cm) v trc honh c ng 2 im chung phn bit. /s: 1m = HT 49. Cho hm s 3 2 2 6 1y x x= + + . Tm m ng thng 1y mx= + ct (C) ti 3 im phn bit A, B, C sao cho A(0;1) v B l trung im ca AC. /s:m = 4 HT 50. Cho hm s 3 21 2 3 3 y x mx x m= + + co ox thi( )mC . Tmm e| ( )mC ca}ttrchoanhtai3ie|mphanbietco to|ng bnhphngcachoanhol nhn15. /s: 1m > HT 51. Cho hm s: 3 2 2 3 1 (1)y x x= + . Tm trn (C) nhng im M sao cho tip tuyn ca (C) ti M ct trc tung ti im c tung bng 8. /s: ( 1; 4)M HT 52. Cho hm s 3 2 2 ( 3) 4y x mx m x= + + + + c th l (Cm) (m l tham s).Cho ng thng (d): 4y x= + v im K(1; 3). Tm cc gi tr ca m (d) ct (Cm) ti ba im phn bit A(0; 4), B, C sao cho tam gic KBC c din tch bng 8 2 . /s: 1 137 2 m = . HT 53. Cho hm s 3 2 3 4y x x= + c th l (C). Gi kd l ng thng i qua im ( 1;0)A vi h s gc k ( )k . Tm k ng thng kd ct th (C) ti ba im phn bit A, B, C v 2 giao im B, C cng vi gc to O to thnh mt tam gic c din tch bng 1 . /s: 1k = HT 54. Cho hm s 3 2 3 2y x x= + c th l (C). Gi E l tm i xng ca th (C). Vit phng trnh ng thng qua E v ct (C) ti ba im E, A, B phn bit sao cho din tch tam gic OAB bng 2 . /s: ( )1; 1 3 ( 1)y x y x= + = . HT 55. Cho hm s 3 24 1 (2 1) ( 2) 3 3 y x m x m x= + + + + c th ( ),mC m l tham s. Gi A l giao im ca 24. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 23 ( )mC vi trc tung. Tm m sao cho tip tuyn ca ( )mC ti A to vi hai trc ta mt tam gic c din tch bng 1 . 3 /s: 13 11 ; 6 6 m m= = HT 56. Cho hm s 3 2y x mx= + + c th (Cm). Tm m th (Cm) ct trc honh ti mt im duy nht. /s: 3m > . HT 57. Cho hm s 3 2 2 3( 1) 6 2y x m x mx= + + c th (Cm).Tm m th (Cm) ct trc honh ti mt im duy nht. /s: 1 3 1 3m < < + HT 58. Cho hm s 3 2 3 1y x x= + . Tm m ng thng (): (2 1) 4 1y m x m= ct th (C) ti ng hai im phn bit. /s: 5 8 m = ; 1 2 m = . HT 59. Cho hm s 3 2 3 ( 1) 1y x mx m x m= + + + c th l ( )mC . Tm tt c cc gi tr ca m : 2 1d y x m= ct th ( )mC ti ba im phn bit c honh ln hn hoc bng 1. /s: khng c gi tr m HT 60. Cho hm s 3 3 2y x x= + (C). Vit phng trnh ng thng ct th (C) ti 3 im phn bit A, B, C sao cho 2Ax = v 2 2BC = /s: : 2d y x= + HT 61. Cho hm s 3 2 4 6 1y x mx= + (C), m l tham s. Tm m ng thng : 1d y x= + ct th hm s ti 3 im A(0;1), B, C vi B, C i xng nhau qua ng phn gic th nht. /s: 2 3 m = HT 62. Cho hm s 3 2 3 1y x x mx= + + + (m l tham s) (1).Tm m ng thng : 1d y = ct th hm s (1) ti ba im phn bit A(0; 1), B, C sao cho cc tip tuyn ca th hm s (1) ti B v C vung gc vi nhau. /s: 9 65 9 65 8 8 m m + = = HT 63. Cho hm s 3 3 1y x x= + c th (C) v ng thng (d): 3y mx m= + + . Tm m (d) ct (C) ti (1; 3)M , N, P sao cho tip tuyn ca (C) ti N v P vung gc vi nhau. /s: 3 2 2 3 2 2 3 3 m m + = = HT 64. Cho hm s 3 2 3 4y x x= + (C). Gi (d) l ng thng i qua im A(2; 0) c h s gc k. Tm k (d) ct (C) ti ba im phn bit A, M, N sao cho hai tip tuyn ca (C) ti M v N vung gc vi nhau. /s: 3 2 2 3 k = HT 65. Cho hm s 3 1 ( ).my x mx m C= + Tm m tip tuyn ca th hm s cho ti im 1x = ct ng trn (C): 2 2 ( 2) ( 3) 4x y + = theo mt dy cung c di nh nht. /s: 2m = HT 66. Cho hm s ( )3 3 2 .my x mx C= + Tm m ng thng i qua im cc i, cc tiu ca( )mC ct ng trn tm ( )1;1 ,I bn knh bng 1 ti hai im phn bit A, B sao cho din tch tam gic IAB t gi tr ln nht /s: 2 3 2 m = HT 67. Cho hm s 4 2 1y x mx m= + c th l ( )mC nh m th ( )mC ct trc trc honh ti bn im www.VNMATH.com 25. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 24 phn bit. /s: 1 2 m m > HT 68. Cho hm s 4 2 2( 1) 2 1 ( ).my x m x m C= + + + Tm tt c cc gi tr ca tham s m th hm s cho ct trc honh ti 4 im phn bit , , ,A B C D ln lt c honh 1 2 3 4, , ,x x x x 1 2 3 4( )x x x x< < < sao cho tam gic ACK c din tch bng 4 bit (3; 2).K /s: 4m = HT 69. Cho hm s ( )4 2 2 1 2 1y x m x m= + + + c th l ( )mC . nh m th ( )mC ct trc honh ti 4 im phn bit c honh lp thnh cp s cng. /s: 4 4; 9 m = HT 70. Cho hm s 4 2 (3 2) 3y x m x m= + + c th l (Cm), m l tham s. Tm m ng thng 1y = ct th (Cm) ti 4 im phn bit u c honh nh hn 2. /s: 1 1 3 0 m m < < HT 71. Cho hm s ( )4 2 2 1 2 1y x m x m= + + + c th l (Cm), m l tham s. Tm m th (Cm) ct trc honh ti 3 im phn bit u c honh nh hn 3. /s: 1 1 2 m m= . HT 72. Cho hm s: 4 2 5 4y x x= + . Tm tt c cc im M trn th (C) ca hm s sao cho tip tuyn ca (C) ti M ct (C) ti hai im phn bit khc M. /s: 10 10 2 2 30 6 m m < < HT 73. Cho hm s 2 1 2 x y x + = + c th l (C). Chng minh rng ng thng :d y x m= + lun ct th (C) ti hai im phn bit A, B. Tm m on AB c di nh nht. /s: 0m = . HT 74. Cho hm s 3 1 x y x = + (C). Vit phng trnh ng thng d qua im ( 1;1)I v ct th (C) ti hai im M, N sao cho I l trung im ca on MN. /s: 1y kx k= + + vi 0k < . HT 75. Cho hm s 2 4 1 x y x + = (C). Gi (d) l ng thng qua A(1; 1) v c h s gc k. Tm k (d) ct (C) ti hai im M, N sao cho 3 10MN = . /s: 3 41 3 41 3; ; 16 16 k k k + = = = HT 76. Cho hm s 2 2 1 x y x = + (C). Tm m ng thng (d): 2y x m= + ct (C) ti hai im phn bit A, B sao cho 5AB = ./s: 10; 2m m= = . HT 77. Cho hm s 1x y x m = + (1). Tm cc gi tr ca tham s m sao cho ng thng (d): 2y x= + ct th hm s (1) ti hai im A v B sao cho 2 2AB = . /s: 7m = HT 78. Cho hm s 2 ( ). 2 2 x y C x + = Tm tt c cc gi tr ca tham s m ng thng :d y x m= + ct th (C) ti hai im phn bit ,A B sao cho 2 2 37 2 OA OB+ = /s: 5 2 2 m m= = 26. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 25 HT 79. Cho hm s ( ). 1 x y C x = Tm tt c cc gi tr ca tham s m ng thng : 1d y mx m= ct th (C) ti hai im phn bit ,A B sao cho 2 2 MA MB+ t gi tr nh nht./s: 1m = HT 80. Cho hm s 1 ( ). 2 x y C x + = Gi d l ng thng qua (2; 0)M v c h s gc l k . Tm k d ct (C) ti hai im phn bit ,A B sao cho : 2MA MB= /s: 2 3 k = HT 81. Cho hm s 3 2 x y x + = + c th (H). Tm m ng thng d :y = 2x + 3m ct (H) ti hai im phn bit sao cho . 4OAOB = vi O l gc ta . /s: 7 12 m = HT 82. Tm trn (H) : 1 2 x y x + = cc im A, B sao cho di on thng AB bng 4 v ng thng AB vung gc vi ng thng .y x= /s: (3 2; 2); (3 2; 2) (3 2; 2); (3 2; 2)+ + A B hoac A B (1 2; 2 2); (1 2; 2 2) (1 2; 2 2); (1 2; 2 2)+ + + + A B hoac A B HT 83. Cho hm s 3 2 x y x + = c th (H). Tm m ng thng : 1d y x m= + + ti hai im phn bit A, B sao cho AOB nhn./s: 3m > HT 84. Cho hm s 3 2 ( ) 2 x y C x + = + . ng thng y x= ct (C) ti hai im A, B. Tm m ng thng y x m= + ct (C) ti hai im C, D sao cho ABCD l hnh bnh hnh. /s: 10m = HT 85. Cho hm s 2 1 1 x y x = (C). Tm m ng thng d: y x m= + ct (C) ti hai im phn bit A, B sao cho OAB vung ti O. /s: 2m = HT 86. Cho hm s 2 1 x m y mx = + (1). Chng minh rng vi mi 0m th hm s (1) ct (d) : 2 2y x m= ti hai im phn bit A, B thuc mt ng (H) c nh. ng thng (d) ct trc Ox, Oy ln lt ti cc im M, N. Tm m 3OAB OMNS S= HT 87. Cho hm s 2 1 ( ). 1 x y C x = Gi I l giao im ca hai tim cn ca (C). Vi gi tr no ca m th ng thng y x m= + ct th (C) ti hai im phn bit A, B v tam gic IAB u. /s: 3 6m = HT 88. Cho hm s ( ) 1 x y C x = . Tm cc gi tr ca m ng thng y x m= + ct th (C) ti hai im phn bit ,A B sao cho ,OA OB bng 0 60 . Vi O l gc ta . /s: 2 6m m= = PHN 4: TIP TUYN HT 89. Cho ham so 2 1 1 x y x = . Vit phng trnh tip tuyn ca (C), bit khong cch t im (1;2)I n tip tuyn bng 2 . /s: 1 0x y+ = v 5 0x y+ = www.VNMATH.com 27. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 26 HT 90. Cho hm s 3 2 (1 2 ) (2 ) 2y x m x m x m= + + + + (1) (m l tham s).Tm tham s m th ca hm s (1) c tip tuyn to vi ng thng d: 7 0x y+ + = gc , bit 1 cos 26 = ./s: 1 4 m hoc 1 2 m HT 91. Cho hm s 3 2 2 ( ).y x x x C= + Tm ta cc im trn trc honh sao cho qua im k c hai tip tuyn vi th (C) v gc gia hai tip tuyn ny bng 0 45 . /s: ;M O 32 ;0 27 M HT 92. Cho hm s 3 2 3 1y x x= + c th (C). Tm hai im A, B thuc th (C) sao cho tip tuyn ca (C) ti A v B song song vi nhau v di on AB = 4 2 . /s: (3;1), ( 1; 3)A B . HT 93. Cho hm s 1 1 x y x + = (C). Tm trn Oy tt c cc im t k c duy nht mt tip tuyn ti (C). /s: (0;1); (0; 1)M M HT 94. Cho hm s 3 3y x x= (C). Tm trn ng thng :d y x= cc im m t k c ng 2 tip tuyn phn bit vi th (C)./s: (2; 2); ( 2;2)A B HT 95. Cho hm s: 3 3 2y x x= + . Tm tt c im trn ng thng 4y = , sao cho t k c ng 2 tip tuyn ti th (C). /s: 2 ( 1;4); ;4 ;(2;4) 3 HT 96. Cho hm s 3 2 3 2y x x= + (C). Tm trn ng thng : 2d y = cc im m t k c 3 tip tuyn phn bit vi th (C). /s: 1 5 3 2 m m m < > HT 97. Cho hm s ( ) ( ) 2 2 1 . 1y x x= + (C). Cho im ( ;0)A a . Tm a t A k c 3 tip tuyn phn bit vi th (C). /s: 3 3 1 1 2 2 < >a hoac a HT 98. Cho hm s 3 2 3 2.y x x= + Tm trn ng thng 2y = cc im m t c th k c 2 tip tuyn ti th hm s v 2 tip tuyn vung gc vi nhau. /s: 1 2; 27 M HT 99. Cho hm s 3 21 ( ) ( 1) (4 3 ) 1 3 y f x mx m x m x= = + + + c th l (Cm). Tm cc gi tr m sao cho trn th (Cm) tn ti mt im duy nht c honh m m tip tuyn ti vung gc vi ng thng : 2 3 0d x y+ = . /s: hay 2 0 3 m m< > . HT 100. Tm tt c cc gi tr m sao cho trn th ( )mC : 3 21 ( 1) (4 3) 1 3 y mx m x m x= + + + tn ti ng hai im c honh dng m tip tuyn ti vung gc vi ng thng (L): 2 3 0x y+ = /s: 1 1 2 0; ; 2 2 3 m HT 101. Cho hm s 2 2 x y x = + (C). Vit phng trnh tip tuyn ca th (C), bit rng khong cch t tm i xng ca th (C) n tip tuyn l ln nht. /s:y x= v 8y x= + . HT 102. Cho hm s 2 2 3 x y x + = + (1). Vit phng trnh tip tuyn ca th hm s (1), bit tip tuyn ct 28. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 27 trc honh, trc tung ln lt ti hai im phn bit A, B v tam gic OAB cn ti gc ta O. /s: 2y x= . HT 103. Cho hm s y = 2 1 1 x x . Lp phng trnh tip tuyn ca th (C) sao cho tip tuyn ny ct cc trc Ox, Oy ln lt ti cc im A v B tho mn OA = 4OB. /s: 1 5 4 4 1 13 4 4 y x y x = + = + . HT 104. Vit phng trnh tip tuyn ca th hm s 2 2 x y x = bit tip tuyn ct Ox, Oy ln lt ti A v B m tam gic OAB tha mn: 2AB OA= /s: 8y x= + HT 105.Cho hm s 2 3 2 x y x = c th (C). Tm trn (C) nhng im M sao cho tip tuyn ti M ca (C) ct hai tim cn ca (C) ti A, B sao cho AB ngn nht. /s: (3;3)M hoc (1;1)M HT 106.Cho hm s 2 3 2 x y x = .Cho M l im bt k trn (C). Tip tuyn ca (C) ti M ct cc ng tim cn ca (C) ti A v B. Gi I l giao im ca cc ng tim cn. Tm to im M sao cho ng trn ngoi tip tam gic IAB c din tch nh nht. /s: (1;1); (3;3)M M HT 107.Cho hm s 3 2 2 2 1 ( ).my x mx m x m C= + + Tm m th hm s tip xc vi trc honh. /s: 3 1 3 2 m m m= = = HT 108.Cho hm s 2 1 1 x y x + = c th (C). Gi I l giao im ca hai tim cn. Tm im M thuc (C) sao cho tip tuyn ca (C) ti M ct 2 tim cn ti A v B vi chu vi tam gic IAB t gi tr nh nht. /s: ( )1 1 3;2 3M + + , ( )2 1 3;2 3M HT 109. Cho hm s: 2 1 x y x + = (C). Cho im (0; )A a . Tm a t A k c 2 tip tuyn ti th (C) sao cho 2 tip im tng ng nm v 2 pha ca trc honh. /s: 2 3 1 a a > . HT 110. Cho hm s y = 2 1 x x + + . Gi I l giao im ca 2 ng tim cn, l mt tip tuyn bt k ca th (C). d l khong cch t I n . Tm gi tr ln nht ca d. /s:GTLN ca d bng 2 khi 0 0 0 2 x x = = HT 111. Cho hm s 2 1 1 x y x + = + . Vit phng trnh tip tuyn ca th (C), bit rng tip tuyn cch u hai im A(2; 4), B(4; 2). /s: 1 5 ; 1; 5 4 4 y x y x y x= + = + = + HT 112. Cho hm s 2 3 2 x y x = (C). Vit phng trnh tip tuyn ti im M thuc (C) bit tip tuyn ct tim cn ng v tim cn ngang ln lt ti A, B sao cho csin gc ABI bng 4 17 , vi I l giao 2 tim cn. /s: Ti 3 0; 2 M : 1 3 4 2 y x= + ; Ti 5 4; 3 M : 1 7 4 2 y x= + www.VNMATH.com 29. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 28 HT 113. Cho hm s 1 2 1 x y x + = (C). Tm gi tr nh nht ca m sao cho tn ti t nht mt im M (C) m tip tuyn ti M ca (C) to vi hai trc ta mt tam gic c trng tm nm trn ng thng 2 1y m= /s: 1 3 m HT 114. Cho hm s 2 1 ( ). 1 x y C x = Tm cc gi tr ca m th hm s (C) tip xc vi ng thng 5.y mx= + /s: 1m = hoc 9m = PHN 5: BIN LUN S NGHIM CA PHNG TRNH HT 115. Cho hm s 3 2 3 1y x x= + + . 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm m phng trnh 3 2 3 2 3 3x x m m = c ba nghim phn bit. /s: {( 1;3) 0;2}m HT 116.Cho hm s 4 2 5 4y x x= + c th (C). 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm m phng trnh 4 2 2| 5 4 | logx x m + = c 6 nghim. Da vo th ta c PT c 6 nghim 9 44 12 9 log 12 144 12 4 m m= = = . HT 117. Cho hm s: 4 2 2 1y x x= + . 1) Kho st s bin thin v v th (C) ca hm s. 2) Bin lun theo m s nghim ca phng trnh: 4 2 22 1 log 0x x m + + = (m> 0) 1 0 2 m< < 1 2 m = 1 1 2 m< < 1m = 1m > 2 nghim 3 nghim 4 nghim 2 nghim v nghim HT 118. Cho hm s 4 2 ( ) 8 9 1y f x x x= = + . 1) Kho st s bin thin v v th (C) ca hm s. 2) Da vo th (C) hy bin lun theo m s nghim ca phng trnh: 4 2 8 cos 9 cos 0x x m + = vi [0; ]x /s: 0m < 0m = 0 1m< < 81 1 32 m < 81 32 m = 81 32 m > 30. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 29 v nghim 1 nghim 2 nghim 4 nghim 2 nghim v nghim HT 119. Cho hm s 1 . 1 x y x + = 1) Kho st s bin thin v v th (C) ca hm s. 2) Bin lun theo m s nghim ca phng trnh 1 . 1 x m x + = 1; 1m m< > 1m = 1 1m < 2 nghim 1 nghim v nghim PHN 6: IM C BIT CA TH HT 120. Cho hm s 3 3 2y x x= + + (C). Tm 2 im trn th hm s sao cho chng i xng nhau qua tm ( 1;3)M . /s:( )1;0 v ( )1;6 HT 121. Cho hm s 3 3 2y x x= + + (C). Tm trn (C) hai im i xng nhau qua ng thng : 2 2 0d x y + = . /s: 7 1 7 7 1 7 ;2 ; ;2 2 2 2 2 2 2 + HT 122. Cho hm s x 3 2 11 3 3 3 x y x= + + . Tm trn th (C) hai im phn bit M, N i xng nhau qua trc tung. /s: 16 16 3; , 3; 3 3 M N . HT 123. Cho hm s 2 1 1 x y x = + (C).Tm im M thuc th (C) tip tuyn ca (C) ti M vi ng thng i qua M v giao im hai ng tim cn c tch cc h s gc bng 9. /s: (0; 3); ( 2;5)M M HT 124. Cho hm s 2 1 1 x y x + = + (C). Tm trn (C) nhng im c tng khong cch n hai tim cn ca (C) nh nht. /s:(0;1);( 2;3) HT 125. Cho hm s 3 4 2 x y x = (C). Tm cc im thuc (C) cch u 2 tim cn. /s: 1 2(1;1); (4;6)M M HT 126. Cho hm s 4 21 1 1 ( ). 4 2 y x x C= + Tm im M thuc (C) sao cho tng khong cch t im M n hai trc ta l nh nht. /s: (0;1)M HT 127. Cho hm s 4 2 0 0 0 02 3 2 1y x x x= + + c th l (C) v ng thng ( ) 2 1x = .Tm trn th (C) im A c khong cch n ( ) l nh nht /s: 1 3 1 ; 3 2 8 A 2 3 1 ; ; 3 2 8 A + www.VNMATH.com 31. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 30 HT 128. Cho hm s 1 2 x y x + = . Tm trn th hm s im M sao cho tng khong cch t M n hai trc ta l nh nht. /s: 1 0; 2 M HT 129. Cho hm s 2 4 1 x y x = + . Tm trn (C) hai im i xng nhau qua ng thng MN bit ( 3;0); ( 1; 1)M N /s:A(0; 4), B(2; 0). HT 130. Cho hm s 2 1 x y x = . Tm trn th (C) hai im B, C thuc hai nhnh sao cho tam gic ABC vung cn ti nh A vi A(2; 0). /s: ( 1;1), (3;3)B C HT 131. Cho hm s 2 1 1 x y x = + . Tm ta im M (C) sao cho khong cch t im ( 1; 2)I ti tip tuyn ca (C) ti M l ln nht. /s: ( )1 3;2 3M + hoc ( )1 3;2 3M + HT 132. Cho hm s 2 2 1 x y x + = . Tm nhng im trn th (C) cch u hai im (2;0), (0;2)A B . /s: 1 5 1 5 1 5 1 5 , ; , 2 2 2 2 + + HT 133. Cho hm s 3 1 x y x = + . Tm trn hai nhnh ca th (C) hai im A v B sao cho AB ngn nht./s: ( ) ( )4 4 4 4 1 4;1 64 , 1 4;1 64A B + + . HT 134. Cho hm s 4 2 2 1y x x= + Tm ta hai im P. Q thuc (C) sao cho ng thng PQ song song vi trc honh v khong cch t im cc i ca (C) n ng thng PQ bng 8 /s:Vy, P(-2;9), Q(2;9) hoc P(2;9); Q(-2;9) HT 135. Cho hm s 2 (3 1) . m x m m y x m + + = + Tm cc im thuc ng thng 1x = m khng c th i qua. /s:Tp hp cc im thuc ng thng 1x = c tung bng a vi a tha mn : 2 10a< < HT 136. Cho hm s 2 1 ( ). 1 x y C x = Tm trn th (C) hai im ,A B phn bit sao cho ba im , , (0; 1)A B I thng hng ng thi tha mn: . 4.IAIB = /s: ( ) ( )2 2;1 2 ; 2 2;1 2A B + + hoc ( ) ( )1 3; 2 3 ; 1 3; 2 3A B + + PHN 7: CC BI TNG HP HT 137.Cho hm s 2 3 ( ). 2 x y C x + = Tm m ng thng : 2d y x m= + ct th ti hai im phn bit sao cho tip tuyn ti hai im ca th hm s song song vi nhau. /s: 2m = HT 138.Cho hm s 3 2 2 2 1 ( ).y x mx mx C= + Tm m th hm s (C) ct trc honh ti 3 im phn bit (1;0),A B v C sao cho 1 2 . 5k k BC+ = trong 1 2,k k ln lt l h s gc tip tuyn ti B, C ca th hm s (C). /s: 1; 2m m= = HT 139.Cho hm s 3 2 3 2 ( ).my x x mx m C= + + Tm m ( )mC ct trc honh ti 3 im phn bit , ,A B C 32. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 31 sao cho tng cc h s gc ca tip tuyn ca ( )mC ti , ,A B C bng 3./s: 2m = www.VNMATH.com 33. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 32 PHN 8: TUYN TP THI I HC T NM 2009 HT 140.(H A 2009) Cho hm s 2 (1) 2 3 x y x + = + . Vit phng trnh tip tuyn ca th hm s (1), bit tip tuyn ct trc honh, trc tung ln lt ti hai im phn bit A, B sao cho tam gic OAB cn ti gc ta O. /s: 2y x= HT 141.(H B 2009) Cho hm s: 4 2 2 4 (1)y x x= .Vi gi tr no ca ,m phng trnh 2 2 2x x m = c ng 6 nghim thc phn bit. /s: 0 1m< < HT 142.(H D 2009) Cho hm s 4 2 (3 2) 3y x m x m= + + c th l ( )mC vi m l tham s. Tm m ng thng 1y = ct th ( )mC ti 4 im phn bit u c honh nh hn 2. /s: 1 1, 0 3 m m < < HT 143.(H A 2010) Cho hm s 3 2 2 (1 ) (1),y x x m x m= + + vi m l tham s thc. Tm m th (1) ct trc honh ti 3 im phn bit c honh 1 2 3, ,x x x tha mn iu kin: 2 2 2 1 2 3 4x x x+ + < /s: 1 1 4 m < < v 0m HT 144. (H B 2010) Cho hm s 2 1 ( ) 1 x y C x + = + . Tm m ng thng 2y x m= + ct th ( )C ti hai im A v B sao cho tam gic OAB c din tch bng 3 (O l gc ta ). /s: 2m = HT 145. (D 2010) Cho hm s 4 2 6 ( )y x x C= + . Vit phng trnh tip tuyn ca th (C) bit tip tuyn vung gc vi ng thng 1 1 6 y x= /s: 6 10y x= + HT 146. (A 2011)Cho hm s 1 ( ) 2 1 x y C x + = . Chng minh rng vi mi m ng thng y x m= + lun ct th ( )C ti hai im phn bit A v B. Gi 1 2,k k ln lt l h s gc ca tip tuyn vi ( )C ti A v B. Tm m tng 1 2k k+ t gi tr ln nht. /s: 1 2k k+ ln nht bng 2 , khi v ch khi 1.m = HT 147. (B 2011) Cho hm s 4 2 2( 1) (1)y x m x m= + + (vi m l tham s). Tm m th hm s (1) c ba im cc tr A, B, C sao cho OA = BC; trong O l gc ta , A l im cc tr thuc trc tung, B v C l hai im cc tr cn li. /s: 2 2 2m = HT 148. (D 2011) Cho hm s 2 1 ( ) 1 x y C x + = + . Tm k ng thng 2 1y kx k= + + ct th ( )C ti hai im phn bit ,A B sao cho khong cch t A v B n trc honh bng nhau. /s: HT 149. (A,A1 2012) Cho hm s 4 2 2 2( 1) (1)y x m x m= + + , vi m l tham s thc. Tm m th hm s (1) c 3 im cc tr to thnh 3 nh ca mt tam gic vung. /s: 0m = HT 150. (B 2012) Cho hm s 3 2 3 3 3 (1),y x mx m= + m l tham s thc. Tm m th hm s (1) c hai im cc tr A v B sao cho tam gic OAB c din tch bng 48. /s: 2m = HT 151. (D 2012) Cho hm s 3 2 22 2 2(3 1) (1), 3 3 y x mx m x m= + l tham s thc. Tm m hm s (1) c hai im cc tr 1 2;x x sao cho: 1 2 1 22( ) 1.x x x x+ + = /s: 2 3 m = HT 152. (A,A1 2013) Cho hm s 3 2 3 3 1 (1)y x x mx= + + , vi m l tham s thc. Tm m hm s (1) 3k = 34. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 33 nghch bin trn khong (0; )+ /s: 1m HT 153. (B 2013) Cho hm s 3 2 2 3( 1) 6 (1),y x m x mx= + + vi m l tham s thc. Tm m th hm s (1) c hai im cc tr A, B sao cho ng thng AB vung gc vi ng thng 2.y x= + /s: 0; 2m m= = HT 154. (D 2013) Cho hm s 3 2 2 3 ( 1) 1 (1),y x mx m x= + + vi m l tham s thc. Tm m ng thng 1y x= + ct th hm s (1) ti ba im phn bit. /s: 8 0; 9 m m< > -----------------------------------------------------------HT----------------------------------------------------------- www.VNMATH.com 35. CHUYN LUYN THI I HC 2013 - 2014 PHNG TRNH M - LOGARIT BIN SON: LU HUY THNG H NI, 8/2013 H V TN: LP :. TRNG : 36. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 1 CHUYN : PHNG TRNH M LOGARIT VN I: LY THA 1. nh ngha lu tha S m C s a Lu tha a * n N = a R . ......n a a a a a = = (n tha s a) 0 = 0a 0 1a a = = * ( )n n N = 0a 1n n a a a = = * ( , ) m m Z n N n = 0a > ( ) m n nm nna a a a b b a = = = = * lim ( , )n n r r Q n N = 0a > lim n r a a = 2. Tnh cht ca lu tha Vi mi a > 0, b > 0 ta c: . . ; ; ( ) ; ( ) . ; a a a a a a a a a ab a b ba b + = = = = = a > 1 : a a > > ; 0 < a < 1 : a a > < Vi 0 < a < b ta c: 0m m a b m< > ; 0m m a b m> < Ch : + Khi xt lu tha vi s m 0 v s m nguyn m th c s a phi khc 0. + Khi xt lu tha vi s m khng nguyn th c s a phi dng. 3. nh ngha v tnh cht ca cn thc Cn bc n ca a l s b sao cho n b a= . Vi a, b 0, m, n N*, p, q Z ta c: .n n n ab a b= ; ( 0) n n n a a b b b = > ; ( ) ( 0) p n np a a a= > ; m n mn a a= ( 0) n mp qp q Neu th a a a n m = = > ; c bit mnn m a a= www.VNMATH.com 37. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 2 Nu n l s nguyn dng l v a < b th n n a b< . Nu n l s nguyn dng chn v 0 < a < b th n n a b< . Ch : + Khi n l, mi s thc a ch c mt cn bc n. K hiu n a . + Khi n chn, mi s thc dng a c ng hai cn bc n l hai s i nhau. 4. Cng thc li kp Gi A l s tin gi, r l li sut mi k, N l s k. S tin thu c (c vn ln li) l: (1 )N C A r= + VN II: LOGARIT 1. nh ngha Vi a > 0, a 1, b > 0 ta c: loga b a b = = Ch : loga b c ngha khi 0, 1 0 a a b > > Logarit thp phn: 10 lg log logb b b= = Logarit t nhin (logarit Nepe): ln loge b b= (vi 1 lim 1 2,718281 n e n = + ) 2. Tnh cht log 1 0a = ; log 1a a = ; log b a a b= ; log ( 0)a b a b b= > Cho a > 0, a 1, b, c > 0. Khi : + Nu a > 1 th log loga a b c b c> > + Nu 0 < a < 1 th log loga a b c b c> < 3. Cc qui tc tnh logarit Vi a > 0, a 1, b, c > 0, ta c: log ( ) log loga a a bc b c= + log log loga a a b b c c = log loga a b b = 4. i c s 38. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 3 Vi a, b, c > 0 v a, b 1, ta c: log log log a b a c c b = hay log .log loga b a b c c= 1 log loga b b a = 1 log log ( 0)aa c c = Bi tp c bn HT 1: Thc hin cc php tnh sau: 1) 2 1 4 log 4.log 2 2) 5 27 1 log .log 9 25 3) 3 loga a 4) 32 log 2log 3 4 9+ 5) 2 2 log 8 6) 9 8 log 2 log 27 27 4+ 7) 3 4 1/3 7 1 log .log log a a a a a a 8) 3 8 6 log 6.log 9.log 2 9) 3 81 2 log 2 4 log 5 9 + 10) 3 9 9 log 5 log 36 4 log 7 81 27 3+ + 11) 75 log 8log 6 25 49+ 12) 2 5 3 log 4 5 13) 6 8 1 1 log 3 log 2 9 4+ 14) 9 2 125 1 log 4 2 log 3 log 27 3 4 5 + + + 15) 36 log 3.log 36 HT 2: So snh cc cp s sau: 1) 4 va log3 1 log 4 3 2) 0,2 va log3 0,1 log 2 0,34 3) 5 2 va log3 4 2 3 log 5 4 4) 1 1 3 2 1 1 log log 80 15 2 va + 5) 13 17 log 150 log 290va 6) va 6 6 1 loglog 3 22 3 HT 3: Tnh gi tr ca biu thc logarit theo cc biu thc cho: 1)Cho 2 log 14 a= . Tnh 49 log 32 theo a. 2)Cho 15 log 3 a= . Tnh 25 log 15 theo a. 3)Cho lg3 0,477= . Tnh lg9000; lg0,000027 ; 81 1 log 100 . 4)Cho 7 log 2 a= . Tnh 1 2 log 28 theo a. HT 4: Tnh gi tr ca biu thc logarit theo cc biu thc cho: 1)Cho 25 log 7 a= ; 2 log 5 b= . Tnh 3 5 49 log 8 theo a, b. www.VNMATH.com 39. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S TI BN Page 4 2)Cho 30 log 3 a= ; 30 log 5 b= . Tnh 30 log 1350 theo a, b. 3)Cho 14 log 7 a= ; 14 log 5 b= . Tnh 35 log 28 theo a, b. 4)Cho 2 log 3 a= ; 3 log 5 b= ; 7 log 2 c= . Tnh 140 log 63 theo a, b, c. VN III: HM S LY THA HM S M HM S LOGARIT 1. Khi nim 1)Hm s lu tha y x = ( l hng s) S m Hm s y x = Tp xc nh D = n (n nguyn dng) n y x= D = R = n (n nguyn m hoc n = 0) n y x= D = R{0} l s thc khng nguyn y x = D = (0; +) Ch : Hm s 1 ny x= khng ng nht vi hm s ( *)n y x n N= . 2)Hm s m x y a= (a > 0, a 1). Tp xc nh: D = R. Tp gi tr: T = (0; +). Khi a > 1 hm s ng bin, khi 0 < a < 1 hm s nghch bin. Nhn trc honh lm tim cn ngang. th: 0 0, a 1) Tp xc nh: D = (0; +). Tp gi tr: T = R. Khi a > 1 hm s ng bin, khi 0 < a < 1 hm s nghch bin. Nhn trc tung lm tim cn ng. th: 2. Gii hn c bit 1 0 1 lim(1 ) lim 1 x x x x x e x + = + = 0 ln(1 ) lim 1 x x x + = 0 1 lim 1 x x e x = 3. o hm ( ) 1 ( 0)x x x = > ; ( ) 1 .u u u = Ch : ( ) 1 01 0 > = n n n vi x neu n chan x vi x neu n len x . ( ) 1 n n n u u n u = ( ) lnx x a a a = ; ( ) ln .u u a a a u = ( )x x e e = ; ( ) .u u e e u = ( ) 1 log lna x x a = ; ( )log lna u u u a = ( ) 1 ln x x = (x > 0); ( )ln u u u = 0 l phng trnh mt cu tm I( a; b; c) v bn knh R = 2 2 2 a b c d+ + . BI TP C BN HT 1. Cho ba vect , ,a b c . Tm m, n ,c a b = : a) ( ) ( ) ( )3; 1; 2 , 1;2; , 5;1;7a b m c= = = b) ( ) ( ) ( )6; 2; , 5; ; 3 , 6;33;10a m b n c= = = HT 2. Xt s ng phng ca ba vect , ,a b c trong mi trng hp sau y: a) ( ) ( ) ( )1; 1;1 , 0;1;2 , 4;2;3a b c= = = b) ( ) ( ) ( )4;3;4 , 2; 1;2 , 1;2;1a b c= = = c) ( ) ( ) ( )3;1; 2 , 1;1;1 , 2;2;1a b c= = = d) ( ) ( ) ( )4;2;5 , 3;1;3 , 2;0;1a b c= = = HT 3. Tm m 3 vect , ,a b c ng phng: a) ( ) ( ) ( )1; ;2 , 1;2;1 , 0; 2;2a m b m c m= = + = b) (2 1;1;2 1); ( 1;2; 2), (2 ; 1;2)a m m b m m c m m= + = + + = + HT 4. Cho cc vect , , ,a b c u . Chng minh ba vect , ,a b c khng ng phng. Biu din vect u theo cc vect , ,a b c : 237. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 5 a) ( ) ( ) ( )2;1;0 , 1; 1;2 , 2;2; 1 (3;7; 7) a b c u = = = = b) ( ) ( ) ( )21; 7;9 , 3; 6;1 , ;1; 7 ( 4;13; 6) a b c u = = = = HT 5. Chng t bn vect , , ,a b c d ng phng: a) ( ) ( ) ( )2; 6;1 , 4; 3; 2 , 4; 2;2 , ( 2; 11;1)a b c d= = = = b) ( ) ( ) ( )2;6; 1 , 2;1; 1 , 4;3;2 , (2;11; 1)a b c d= = = = HT 6. Cho ba vect , ,a b c khng ng phng v vect d . Chng minh b ba vect sau khng ng phng: a) , ,b c d ma nb= + (vi m, n 0) b) , ,a c d ma nb= + (vi m, n 0) HT 7. Cho im M. Tm ta hnh chiu vung gc ca im M: Trn cc mt phng ta : Oxy, Oxz, Oyz Trn cc trc ta : Ox, Oy, Oz a) (1;2;3)M b) (3; 1;2)M c) ( 1;1; 3)M d) (1;2; 1)M HT 8. Cho im M. Tm ta ca im M i xng vi im M: Qua gc to Qua mp(Oxy) Qua trc Oy a) (1;2;3)M b) (3; 1;2)M c) ( 1;1; 3)M d) (1;2; 1)M HT 9. Xt tnh thng hng ca cc b ba im sau: a) (1;3;1), (0;1;2), (0;0;1)A B C b) (1;1;1), ( 4;3;1), ( 9;5;1)A B C HT 10. Cho ba im A, B, C. Chng t ba im A, B, C to thnh mt tam gic. Tm to trng tm G ca ABC. Xc nh im D sao cho ABCD l hnh bnh hnh. a) (1;2; 3), (0;3;7), (12;5;0)A B C b) (0;13;21), (11; 23;17), (1;0;19)A B C c) (3; 4;7), ( 5;3; 2), (1;2; 3)A B C d) (4;2;3), ( 2;1; 1), (3;8;7)A B C HT 11. Trn trc Oy (Ox), tm im cch u hai im: a) (3;1;0)A , ( 2;4;1)B b) (1; 2;1), (11;0;7)A B c) (4;1;4), (0;7; 4)A B HT 12. Trn mt phng Oxy (Oxz, Oyz), tm im cch u ba im: a) (1;1;1), ( 1;1;0), (3;1; 1)A B C b) ( 3;2;4), (0;0;7), ( 5;3;3)A B C HT 13. Cho hai im A, B. ng thng AB ct mt phng Oyz (Oxz, Oxy) ti im M. im M chia on thng AB theo t s no ? Tm ta im M. a) ( ) ( )2; 1;7 , 4;5; 2A B b) (4;3; 2), (2; 1;1)A B c) (10;9;12), ( 20;3;4)A B HT 14. Cho bn im A, B, C, D. Chng minh A, B, C, D l bn nh ca mt t din. Tm ta trng tm G ca t din ABCD. Tnh gc to bi cc cnh i din ca t din ABCD. Tnh th tch ca khi t din ABCD. Tnh din tch tam gic BCD, t suy ra di ng cao ca t din v t A. 238. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 6 a) (2;5; 3), (1;0;0), (3;0; 2), ( 3; 1;2)A B C D b) ( ) ( ) ( ) ( )1;0;0 , 0;1;0 , 0;0;1 , 2;1; 1A B C D c) ( ) ( ) ( ) ( )1;1;0 , 0;2;1 , 1;0;2 , 1;1;1A B C D d) ( ) ( ) ( ) ( )2;0;0 , 0;4;0 , 0;0;6 , 2;4;6A B C D HT 15. Cho hnh hp ABCD.A'B'C'D'. Tm to cc nh cn li. Tnh th tch khi hp. a) ( ) ( ) ( ) ( )1;0;1 , 2;1;2 , 1; 1;1 , ' 4;5; 5A B D C b) 2 5 3 1 0 0 3 0 2 3 1 2A B C A( ; ; ), ( ; ; ), ( ; ; ), '( ; ; ) c) (0;2;1), (1; 1;1), (0;0;0;), '( 1;1;0)A B D A d) (0;2;2), (0;1;2), ( 1;1;1), '(1; 2; 1)A B C C HT 16. Cho bn im S(3; 1; 2), A(5; 3; 1), B(2; 3; 4), C(1; 2; 0). a) Chng minh SA (SBC), SB (SAC), SC (SAB). b) Chng minh S.ABC l mt hnh chp u. c) Xc nh to chn ng cao H ca hnh chp. Suy ra di ng cao SH. ----------------------------------------------------------------- 239. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 7 BI 2: PHNG TRNH MT CU vit phng trnh mt cu (S), ta cn xc nh tm I v bn knh R ca mt cu. Dng 1: (S) c tm I(a; b; c) v bn knh R: (S): 2 2 2 2 ( ) ( ) ( )x a y b z c R + + = Dng 2: (S) c tm I(a; b; c) v i qua im A: Khi bn knh R = IA. Dng 3: (S) nhn on thng AB cho trc lm ng knh: Tm I l trung im ca on thng AB: ; ; 2 2 2 A B A B A B I I I x x y y z z x y z + + + = = = . Bn knh R = IA = 2 AB . Dng 4: (S) i qua bn im A, B, C, D (mt cu ngoi tip t din ABCD): Gi s phng trnh mt cu (S) c dng: 2 2 2 2 2 2 0x y z ax by cz d+ + + + + + = (*). Thay ln lt to ca cc im A, B, C, D vo (*), ta c 4 phng trnh. Gii h phng trnh , ta tm c a, b, c, d Phng trnh mt cu (S). Dng 5: (S) i qua ba im A, B, C v c tm I nm trn mt phng (P) cho trc: Gii tng t nh dng 4. Dng 6: (S) c tm I v tip xc vi mt cu (T) cho trc: Xc nh tm J v bn knh R ca mt cu (T). S dng iu kin tip xc ca hai mt cu tnh bn knh R ca mt cu (S). (Xt hai trng hp tip xc trong v tip xc ngoi) Ch : Vi phng trnh mt cu (S): 2 2 2 2 2 2 0x y z ax by cz d+ + + + + + = vi 2 2 2 0a b c d+ + > th (S) c tm I(a; b; c) v bn knh R = 2 2 2 a b c d+ + . BI TP C BN HT 17. Tm tm v bn knh ca cc mt cu sau: a) 2 2 2 8 2 1 0x y z x y+ + + + = b) 2 2 2 4 8 2 4 0x y z x y z+ + + + = c) 2 2 2 2 4 4 0x y z x y z+ + + = d) 2 2 2 6 4 2 86 0x y z x y z+ + + = HT 18. Vit phng trnh mt cu c tm I v bn knh R: a) (1; 3;5), 3I R = b) (5; 3;7), 2I R = c) (1; 3;2), 5I R = d) (2;4; 3), 3I R = HT 19. Vit phng trnh mt cu c tm I v i qua im A: a) (2;4; 1), (5;2;3)I A b) (0;3; 2), (0;0;0)I A c) (3; 2;1), (2;1; 3)I A HT 20. Vit phng trnh mt cu c ng knh AB, vi: a) (2;4; 1), (5;2;3)A B b) (0;3; 2), (2;4; 1)A B c) (3; 2;1), (2;1; 3)A B 240. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 8 HT 21. Vit phng trnh mt cu ngoi tip t din ABCD, vi: a) ( ) ( ) ( ) ( )1;1;0 , 0;2;1 , 1;0;2 , 1;1;1A B C D b) ( ) ( ) ( ) ( )2;0;0 , 0;4;0 , 0;0;6 , 2;4;6A B C D HT 22. Vit phng trnh mt cu i qua ba im A, B, C v c tm nm trong mt phng (P) cho trc, vi: a) (1;2;0), ( 1;1;3), (2;0; 1) ( ) ( ) A B C P Oxz b) (2;0;1), (1;3;2), (3;2;0) ( ) ( ) A B C P Oxy HT 23. Vit phng trnh mt cu (S) c tm I v tip xc vi mt cu (T), vi: a) 2 2 2 ( 5;1;1) ( ) : 2 4 6 5 0 I T x y z x y z + + + + = b) 2 2 2 ( 3;2;2) ( ) : 2 4 8 5 0 I T x y z x y z + + + + = -------------------------------------------------------------------- BI 3: PHNG TRNH MT PHNG 1. Vect php tuyn Cp vect ch phng ca mt phng Vect 0n l VTPT ca () nu gi ca n vung gc vi (). Ch : Nu n l mt VTPT ca () th kn (k 0) cng l VTPT ca (). 2. Phng trnh tng qut ca mt phng 2 2 2 0 0Ax By Cz D vi A B C+ + + = + + > Nu () c phng trnh 0Ax By Cz D+ + + = th ( ; ; )n A B C= l mt VTPT ca (). Phng trnh mt phng i qua 0 0 0 0( ; ; )M x y z v c mt VTPT ( ; ; )n A B C= l: 0 0 0( ) ( ) ( ) 0A x x B y y C z z + + = 3. Cc trng hp ring Ch : Nu trong phng trnh ca () khng cha n no th () song song hoc chatrc tng ng. Phng trnh mt phng theo on chn: 1 x y z a b c + + = () ct cc trc to ti cc im (a; 0; 0), (0; b; 0), (0; 0; c) 4. V tr tng i ca hai mt phng Cho hai mt phng (), () c phng trnh: (): 1 1 1 1 0A x B y C z D+ + + = (): 2 2 2 2 0A x B y C z D+ + + = Cc h s Phng trnh mt phng () Tnh cht mt phng () D = 0 () i qua gc to O A = 0 () // Ox hoc () Ox B = 0 () // Oy hoc () Oy C = 0 () // Oz hoc () Oz A = B = 0 () // (Oxy) hoc () (Oxy) A = C = 0 () // (Oxz) hoc () (Oxz) B = C = 0 () // (Oyz) hoc () (Oyz) 241. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 9 (), () ct nhau 1 1 1 2 2 2: : : :A B C A B C () // () 1 1 1 1 2 2 2 2 A B C D A B C D = = () () 1 1 1 1 2 2 2 2 A B C D A B C D = = = () () 1 2 1 2 1 2 0A A B B C C+ + = 5. Khong cch t im M0(x0; y0; z0) n mt phng (): Ax + By + Cz + D = 0 ( ) 0 0 0 0 2 2 2 ,( ) Ax By Cz D d M A B C + + + = + + VN 1: Vit phng trnh mt phng lp phng trnh mt phng () ta cn xc nh mt im thuc () v mt VTPT ca n. Dng 1: () i qua im ( )0 0 0; ;M x y z c VTPT ( ); ;n A B C= : (): ( ) ( ) ( )0 0 0 0A x x B y y C z z + + = Dng 2: () i qua im ( )0 0 0; ;M x y z c cp VTCP ,a b : Khi mt VTPT ca () l ,n a b = . Dng 3: () i qua im ( )0 0 0; ;M x y z v song song vi mt phng (): Ax + By + Cz + D = 0: (): ( ) ( ) ( )0 0 0 0A x x B y y C z z + + = Dng 4: () i qua 3 im khng thng hng A, B, C: Khi ta c th xc nh mt VTPT ca () l: ,n AB AC = Dng 5: () i qua mt im M v mt ng thng (d) khng cha M: Trn (d) ly im A v VTCP u . Mt VTPT ca () l: ,n AM u = Dng 6: () i qua mt im M v vung gc vi mt ng thng (d): VTCP u ca ng thng (d) l mt VTPT ca (). Dng 7: () i qua 2 ng thng ct nhau d1, d2: Xc nh cc VTCP ,a b ca cc ng thng d1, d2. Mt VTPT ca () l: ,n a b = . Ly mt im M thuc d1 hoc d2 M (). Dng 8: () cha ng thng d1 v song song vi ng thng d2 (d1, d2 cho nhau): Xc nh cc VTCP ,a b ca cc ng thng d1, d2. 242. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 10 Mt VTPT ca () l: ,n a b = . Ly mt im M thuc d1 M (). Dng 9: () i qua im M v song song vi hai ng thng cho nhau d1, d2: Xc nh cc VTCP ,a b ca cc ng thng d1, d2. Mt VTPT ca () l: ,n a b = . Dng 10: () i qua mt ng thng (d) v vung gc vi mt mt phng (): Xc nh VTCP u ca (d) v VTPT n ca (). Mt VTPT ca () l: ,n u n = . Ly mt im M thuc d M (). Dng 11: () i qua im M v vung gc vi hai mt phng ct nhau (), (): Xc nh cc VTPT ,n n ca () v (). Mt VTPT ca () l: ,n u n = . Dng 12: () i qua ng thng (d) cho trc v cch im M cho trc mt khong k cho trc: Gi s () c phng trnh: Ax z+D 0By C+ + = ( )2 2 2 0A B C+ + . Ly 2 im A, B (d) A, B () (ta c hai phng trnh (1), (2)). T iu kin khong cch ( ,( ))d M k = , ta c phng trnh (3). Gii h phng trnh (1), (2), (3) (bng cch cho gi tr mt n, tm cc n cn li). Dng 13: () l tip xc vi mt cu (S) ti im H: Gi s mt cu (S) c tm I v bn knh R. Mt VTPT ca () l: n IH= Ch : vit phng trnh mt phng cn nm vng cc cch xc nh mt phng hc lp 11. BI TP C BN HT 24. Vit phng trnh mt phng (P) i qua im M v c VTPT cho trc: a) ( ) ( )3;1;1 , 1;1;2M n = b) ( ) ( )2;7;0 , 3;0;1M n = c) ( ) ( )4; 1; 2 , 0;1;3M n = HT 25. Vit phng trnh mt phng trung trc ca on thng AB cho trc, vi: a) (2;1;1), (2; 1; 1)A B b) (1; 1; 4), (2;0;5)A B c) (2;3; 4), (4; 1;0)A B HT 26. Vit phng trnh mt phng i qua im M v c cp VTCP ,a b cho trc, vi: a) (1;2; 3), (2;1;2), (3;2; 1)M a b = = b) (1; 2;3), 3; 1; 2), (0;3;4)M a b = = HT 27. Vit phng trnh mt phng () i qua im M v song song vi mt phng ( ) cho trc, vi: a) ( ) ( ) ( )2;1;5 ,M Oxy = b) ( ) ( )1; 2;1 , : 2 3 0M x y + = HT 28. Vit phng trnh mt phng () i qua im M v ln lt song song vi cc mt phng to , vi: 243. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 11 a) ( )2;1;5M b) ( )1; 2;1M c) ( )1;1;0M d) ( )3;6; 5M HT 29. Vit phng trnh mt phng () i qua ba im A, B, C khng thng hng cho trc, vi: a) (1; 2;4), (3;2; 1), ( 2;1; 3)A B C b) (0;0;0), ( 2; 1;3), (4; 2;1)A B C HT 30. Vit phng trnh mt phng () i qua im A v vung gc vi ng thng i qua hai im B, C cho trc, vi: a) (1; 2;4), (3;2; 1), ( 2;1; 3)A B C b) (0;0;0), ( 2; 1;3), (4; 2;1)A B C HT 31. Vit phng trnh mt phng () i qua hai im A, B v vung gc vi mt phng () cho trc, vi: a) ( ) (3;1; 1), (2; 1;4) : 2 3 1 0 A B x y z + = b) ( ) ( 2; 1;3), (4; 2;1) : 2 3 2 5 0 A B x y z + + = c) ( ) (2; 1;3), ( 4;7; 9) : 3 4 8 5 0 A B x y z + = HT 32. Vit phng trnh mt phng () i qua im M v vung gc vi hai mt phng (), () cho trc, vi: a) ( ) ( )( 1; 2;5), : 2 3 1 0, : 2 3 1 0M x y z x y z + + = + + = b) ( ) ( )(1;0; 2), : 2 2 0, : 3 0M x y z x y z + = = HT 33. Vit phng trnh mt phng () i qua im M v giao tuyn ca hai mt phng (P), (Q) cho trc, vi: a) ( ) ( ) ( ):1;2; 3 , : 2 3 5 0, 3 2 5 1 0M P x y z Q x y z + = + = b) ( ) ( ) ( ):2;1; 1 , : 4 0, 3 1 0M P x y z Q x y z + = + = HT 34. Vit phng trnh mt phng () qua giao tuyn ca hai mt phng (P), (Q), ng thi song song vi mt phng (R) cho trc, vi: a) ( ) : 2 4 0, ( ) : 3 0, ( ) : 2 0P y z Q x y z R x y z+ = + = + + = b) ( ) : 4 2 5 0, ( ) : 4 5 0, ( ) : 2 19 0P x y z Q y z R x y + = + = + = c) ( ) : 3 2 0, ( ) : 4 5 0, ( ) : 2 7 0P x y z Q x y R x z + = + = + = HT 35. Vit phng trnh mt phng () qua giao tuyn ca hai mt phng (P), (Q), ng thi vung gc vi mt phng (R) cho trc, vi: a) ( ) : 2 3 4 0, ( ) : 2 3 5 0, ( ) : 2 3 2 0P x y Q y z R x y z+ = = + = b) ( ) : 2 4 0, ( ) : 3 0, ( ) : 2 0P y z Q x y z R x y z+ = + + = + + = c) ( ) : 2 4 0, ( ) : 2 5 0, ( ) : 2 3 6 0P x y z Q x y z R x y z+ = + + + = + = d) ( ) : 3 2 0, ( ) : 4 5 0, ( ) : 2 7 0P x y z Q x y R x z + = + = + = HT 36. Vit phng trnh mt phng () qua giao tuyn ca hai mt phng (P), (Q), ng thi cch im M cho trc mt khong bng k, vi: a) ( ): 2 0, ( ) : 5 13 2 0, (1;2;3), 2P x y Q x y z M k = + = = VN 2: V tr tng i ca hai mt phng HT 37. Xt v tr tng i ca cc cp mt phng sau: a) 2 3 2 5 0 3 4 8 5 0 x y z x y z + + = + = b) 3 4 3 6 0 3 2 5 3 0 x y z x y z + + = + = c) 5 5 5 1 0 3 3 3 7 0 x y z x y z + = + + = HT 38. Xc nh m, n cc cp mt phng sau: song song ct nhau trng nhau a) 3 2 7 0 7 6 4 0 x my z nx y z + = + + = b) 5 2 11 0 3 5 0 x y mz x ny z + = + + = c) 2 3 5 0 6 6 2 0 x my z nx y z + + = + = 244. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 12 HT 39. Xc nh m cc cp mt phng sau vung gc vi nhau a) 2 7 2 0 3 2 15 0 x y mz x y z + + = + + = b) (2 1) 3 2 3 0 ( 1) 4 5 0 m x my z mx m y z + + = + + = VN 3: Khong cch t mt im n mt mt phng. Khong cch gia hai mt phng song song. Hnh chiu ca mt im trn mt phng . im i xng ca mt im qua mt phng. Khong cch t im M0(x0; y0; z0) n mt phng (): Ax + By + Cz + D = 0 ( ) 0 0 0 0 2 2 2 ,( ) Ax By Cz D d M A B C + + + = + + Khong cch gia hai mt phng song song bng khong cch t mt im bt k trn mt phng ny n mt phng kia. Ch : Nu hai mt phng khng song song th khong cch gia chng bng 0. im H l hnh chiu ca im M trn (P) , ( ) MH n cung phng H P im M i xng vi im M qua (P) 2MM MH = BI TP HT 40. Cho mt phng (P) v im M. Tnh khong cch t M n (P). Tm to hnh chiu H ca M trn (P). Tm to im M i xng vi M qua (P). a) ( ) : 2 2 6 0, (2; 3;5)P x y z M + = b) ( ) : 5 14 0, (1; 4; 2)P x y z M+ + = HT 41. Tm khong cch gia hai mt phng: a) 2 3 1 0 2 3 5 0 x y z x y z + + = + + = b) 6 2 1 0 6 2 3 0 x y z x y z + + = + = c) 2 4 5 0 3 5 1 0 x y z x y z + + = + = HT 42. Tm im M trn trc Ox (Oy, Oz) cch u im N v mt phng (P): a) ( ) : 2 2 5 0, (1;2; 2)P x y z N+ + = b) ( ) : 5 14 0, (1; 4; 2)P x y z N+ + = c) ( ) : 6 2 3 12 0, (3;1; 2)P x y z N + + = d) ( ) : 2 4 4 3 0, (2; 3;4)P x y z N + + = HT 43. Tm im M trn trc Ox (Oy, Oz) cch u hai mt phng: a) 1 0 5 0 x y z x y z + + = + = b) 2 2 1 0 2 2 5 0 x y z x y z + + = + + = c) 2 4 5 0 4 2 1 0 x y z x y z + + = + = HT 44. Tm phng trnh tng qut ca mt phng (P) i qua im A v song song vi mt phng (Q) cho trc. Tnh khong cch gia (P) v (Q): a) ( )1;2;3 , ( ) : 2 4 4 0A Q x y z + = . b) ( )3; 1;2 , ( ) : 6 2 3 12 0A Q x y z + + = . HT 45. Tm phng trnh tng qut ca mt phng (P) song song vi mt phng (Q) v cch im A mt khong k cho trc: a) ( ) : 2 2 5 0, (2; 1;4), 4Q x y z A k+ + = = b) ( ) : 2 4 4 3 0, (2; 3;4), 3Q x y z A k + + = = HT 46. Tm phng trnh tng qut ca mt phng (P) cch mt phng (Q) mt khong k: 245. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 13 a) ( ) : 3 2 3 0, 14Q x y z k + = = b) ( ) : 4 3 2 5 0, 29Q x y z k+ + = = VN 4: Gc gia hai mt phng Cho hai mt phng (), () c phng trnh: (): 1 1 1 1 0A x B y C z D+ + + = (): 2 2 2 2 0A x B y C z D+ + + = Gc gia (), () bng hoc b vi gc gia hai VTPT 1 2,n n . ( ) 1 2 1 2 1 2 1 2 2 2 2 2 2 2 1 2 1 1 1 2 2 2 . cos ( ),( ) . . n n A A B B C C n n A B C A B C + + = = + + + + Ch : ( )0 0 0 ( ),( ) 90 . 1 2 1 2 1 2( ) ( ) 0A A B B C C + + = BI TP C BN HT 47. Tnh gc gia hai mt phng: a) 1 0 5 0 x y z x y z + + = + = b) 2 2 1 0 2 2 5 0 x y z x y z + + = + + = c) 2 4 5 0 4 2 1 0 x y z x y z + + = + = d) 4 4 2 7 0 2 4 5 0 x y z x z + + = + = e) 2 2 3 0 2 2 12 0 x y z y z + = + + = f) 3 3 3 2 0 4 2 4 9 0 x y z x y z + + = + + = HT 48. Tm m gc gia hai mt phng sau bng cho trc: a) 0 (2 1) 3 2 3 0 ( 1) 4 5 0 90 m x my z mx m y z + + = + + = = b) 0 2 12 0 7 0 45 mx y mz x my z + + = + + + = = ----------------------------------------------------------------- BI 4: PHNG TRNH NG THNG 1. Phng trnh tham s ca ng thng Phng trnh tham s ca ng thng d i qua im 0 0 0 0( ; ; )M x y z v c VTCP 1 2 3( ; ; )a a a a= : 1 2 3 ( ) : ( ) o o o x x a t d y y a t t R z z a t = + = + = + Nu 1 2 3 0a a a th 0 0 0 1 2 3 ( ) : x x y y z z d a a a = = c gi l phng trnh chnh tc ca d. 2. V tr tng i gia hai ng thng Cho hai ng thng d, d c phng trnh tham s ln lt l: 0 1 0 2 0 3 : x x ta d y y ta z z ta = + = + = + v 0 1 0 2 0 3 : x x t a d y y t a z z t a = + = + = + 246. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 14 d // d 0 1 0 1 0 2 0 2 0 3 0 3 , ( , ) + = + + = + + = + a a cung phng x ta x t a he y ta y t a an t t vo nghiem z ta z t a 0 0 0 0 , ( ; ; ) a a cung phng M x y z d 0 0 , , a a cung phng a M M khong cung phng 0 0 , 0 , 0 = a a a M M d d 0 1 0 1 0 2 0 2 0 3 0 3 ( , ) + = + + = + + = + x ta x t a he y ta y t a an t t co vo so nghiem z ta z t a 0 0 0 0 , ( ; ; ) a a cung phng M x y z d 0 0, , a a M M oi mot cung phng 0 0, , 0 = = a a a M M d, d ct nhau h 0 1 0 1 0 2 0 2 0 3 0 3 + = + + = + + = + x ta x t a y ta y t a z ta z t a (n t, t) c ng mt nghim 0 0 , , , a a khong cung phng a a M M ong phang 0 0 , 0 , . 0 = a a a a M M d, d cho nhau 0 1 0 1 0 2 0 2 0 3 0 3 , ( , ) + = + + = + + = + a a khong cung phng x ta x t a he y ta y t a an t t vo nghiem z ta z t a 0 0, , a a M M khong ong phang 0 0, . 0 a a M M d d a a . 0a a = 3. V tr tng i gia mt ng thng v mt mt phng Cho mt phng (): 0Ax By Cz D+ + + = v ng thng d: 0 1 0 2 0 3 x x ta y y ta z z ta = + = + = + Xt phng trnh: 0 1 0 2 0 3( ) ( ) ( ) 0A x ta B y ta C z ta D+ + + + + + = (n t) (*) d // () (*) v nghim d ct () (*) c ng mt nghim d () (*) c v s nghim 4. V tr tng i gia mt ng thng v mt mt cu 247. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 15 Cho ng thng d: 0 1 0 2 0 3 x x ta y y ta z z ta = + = + = + (1) v mt cu (S): 2 2 2 2 ( ) ( ) ( )x a y b z c R + + = (2) xt VTT ca d v (S) ta thay (1) vo (2), c mt phng trnh (*). d v (S) khng c im chung (*) v nghim d(I, d) > R d tip xc vi (S) (*) c ng mt nghim d(I, d) = R d ct (S) ti hai im phn bit (*) c hai nghim phn bit d(I, d) < R 5. Khong cch t mt im n mt ng thng (chng trnh nng cao) Cho ng thng d i qua M0 v c VTCP a v im M. 0 , ( , ) M M a d M d a = 6. Khong cch gia hai ng thng cho nhau (chng trnh nng cao) Cho hai ng thng cho nhau d1 v d2. d1 i qua im M1 v c VTCP 1a , d2 i qua im M2 v c VTCP 2a 1 2 1 2 1 2 1 2 , . ( , ) , a a M M d d d a a = Ch : Khong cch gia hai ng thng cho nhau d1, d2 bng khong cch gia d1 vi mt phng () cha d2 v song song vi d1. 7. Khong cch gia mt ng thng v mt mt phng song song Khong cch gia ng thng d vi mt phng () song song vi n bng khong cch t mt im M bt k trn d n mt phng (). 8. Gc gia hai ng thng Cho hai ng thng d1, d2 ln lt c cc VTCP 1 2,a a . Gc gia d1, d2 bng hoc b vi gc gia 1 2,a a . ( ) 1 2 1 2 1 2 . cos , . a a a a a a = 9. Gc gia mt ng thng v mt mt phng Cho ng thng d c VTCP 1 2 3( ; ; )a a a a= v mt phng () c VTPT ( ; ; )n A B C= . Gc gia ng thng d v mt phng () bng gc gia ng thng d vi hnh chiu d ca n trn (). ( ) 1 2 3 2 2 2 2 2 2 1 2 3 sin ,( ) . Aa Ba Ca d A B C a a a + + = + + + + 248. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 16 VN 1: Lp phng trnh ng thng lp phng trnh ng thng d ta cn xc nh mt im thuc d v mt VTCP ca n. Dng 1: d i qua im 0 0 0 0( ; ; )M x y z v c VTCP 1 2 3( ; ; )a a a a= : 1 2 3 ( ) : ( ) o o o x x a t d y y a t t R z z a t = + = + = + Dng 2: d i qua hai im A, B: Mt VTCP ca d l AB . Dng 3: d i qua im 0 0 0 0( ; ; )M x y z v song song vi ng thng cho trc: V d // nn VTCP ca cng l VTCP ca d. Dng 4: d i qua im 0 0 0 0( ; ; )M x y z v vung gc vi mt phng (P) cho trc: V d (P) nn VTPT ca (P) cng l VTCP ca d. Dng 5: d l giao tuyn ca hai mt phng (P), (Q): Cch 1: Tm mt im v mt VTCP. Tm to mt im A d: bng cch gii h phng trnh ( ) ( ) P Q (vi vic chn gi tr cho mt n) Tm mt VTCP ca d: ,P Qa n n = Cch 2: Tm hai im A, B thuc d, ri vit phng trnh ng thng i qua hai im . Dng 6: d i qua im 0 0 0 0( ; ; )M x y z v vung gc vi hai ng thng d1, d2: V d d1, d d2 nn mt VTCP ca d l: 1 2 ,d da a a = Dng 7: d i qua im 0 0 0 0( ; ; )M x y z , vung gc v ct ng thng . Cch 1: Gi H l hnh chiu vung gc ca M0 trn ng thng . 0 H M H u Khi ng thng d l ng thng i qua M0, H. Cch 2: Gi (P) l mt phng i qua A v vung gc vi d; (Q) l mt phng i qua A v cha d. Khi d = (P) (Q) Dng 8: d i qua im 0 0 0 0( ; ; )M x y z v ct hai ng thng d1, d2: Cch 1: Gi M1 d1, M2 d2. T iu kin M, M1, M2 thng hng ta tm c M1, M2. T suy ra phng trnh ng thng d. Cch 2: Gi (P) = 0 1( , )M d , (Q) = 0 2( , )M d . Khi d = (P) (Q). Do , mt VTCP ca d c th chn l ,P Qa n n = . Dng 9: d nm trong mt phng (P) v ct c hai ng thng d1, d2: 249. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 17 Tm cc giao im A = d1 (P), B = d2 (P). Khi d chnh l ng thng AB. Dng 10: d song song vi v ct c hai ng thng d1, d2: Vit phng trnh mt phng (P) cha v d1, mt phng (Q) cha v d2. Khi d = (P) (Q). Dng 11: d l ng vung gc chung ca hai ng thng d1, d2 cho nhau: Cch 1: Gi M d1, N d2. T iu kin 1 2 MN d MN d , ta tm c M, N. Khi , d l ng thng MN. Cch 2: V d d1 v d d2 nn mt VTCP ca d c th l: 1 2 ,d da a a = . Lp phng trnh mt phng (P) cha d v d1, bng cch: + Ly mt im A trn d1. + Mt VTPT ca (P) c th l: 1 ,P dn a a = . Tng t lp phng trnh mt phng (Q) cha d v d2. Khi d = (P) (Q). Dng 12: d l hnh chiu ca ng thng ln mt phng (P): Lp phng trnh mt phng (Q) cha v vung gc vi mt phng (P) bng cch: Ly M . V (Q) cha v vung gc vi (P) nn ,Q Pn a n = . Khi d = (P) (Q). Dng 13: d i qua im M, vung gc vi d1 v ct d2: Cch 1: Gi N l giao im ca d v d2. T iu kin MN d1, ta tm c N. Khi , d l ng thng MN. Cch 2: Vit phng trnh mt phng (P) qua M v vung gc vi d1. Vit phng trnh mt phng (Q) cha M v d2. Khi d = (P) (Q). BI TP C BN HT 49. Vit phng trnh tham s ca ng thng i qua im M v c VTCP a cho trc: a) (1;2; 3), ( 1;3;5)M a = b) (0; 2;5), (0;1;4)M a = c) (1;3; 1), (1;2; 1)M a = d) (3; 1; 3), (1; 2;0)M a = e) (3; 2;5), ( 2;0;4)M a = f) (4;3; 2), ( 3;0;0)M a = HT 50. Vit phng trnh tham s ca ng thng i qua hai im A, B cho trc: 250. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 18 a) ( ) ( ),2;3; 1 1;2;4A B b) ( ) ( ),1; 1;0 0;1;2A B c) ( ) ( ),3;1; 5 2;1; 1A B d) ( ) ( ),2;1;0 0;1;2A B e) ( ) ( )A ,1;2; 7 1;2;4B f) ( ) ( ),2;1;3 4;2; 2A B HT 51. Vit phng trnh tham s ca ng thng i qua im A v song song vi ng thng cho trc: a) ( ),3;2; 4A Ox b) ( )2; 5;3 , (5;3;2), (2;1; 2)A qua M N c) 2 3 (2; 5;3), : 3 4 5 2 x t A y t z t = = + = d) 2 5 2 (4; 2;2), : 4 2 3 x y z A + = = HT 52. Vit phng trnh tham s ca ng thng i qua im A v vung gc vi mt phng (P) cho trc: a) ( ), (P)2;4;3 : 2 3 6 19 0A x y z + + = b) ( ),1; 1;0 ( ) : ( )A P Oxy c) ( )3;2;1 , ( ) : 2 5 4 0A P x y + = d) (2; 3;6), ( ) : 2 3 6 19 0A P x y z + + = HT 53. Vit phng trnh tham s ca ng thng l giao tuyn ca hai mt phng (P), (Q) cho trc: a) ( ) : 6 2 2 3 0 ( ) : 3 5 2 1 0 P x y z Q x y z + + + = = b) ( ) : 2 3 3 4 0 ( ) : 2 3 0 P x y z Q x y z + = + + = c) ( ) : 3 3 4 7 0 ( ) : 6 2 6 0 P x y z Q x y z + + = + + = d) ( ) : 2 3 0 ( ) : 1 0 P x y z Q x y z + + = + + = e) ( ) : 1 0 ( ) : 2 0 P x z Q y + = = f) ( ) : 2 1 0 ( ) : 1 0 P x y z Q x z + + = + = HT 54. Vit phng trnh tham s ca ng thng i qua im A v vung gc vi hai ng thng d1, d2 cho trc: a) 1 2 1 2 1 (1;0;5), : 3 2 , : 2 1 1 3 x t x t A d y t d y t z t z t = + = = = + = + = b) 1 2 1 1 3 (2; 1;1), : 2 , : 2 3 3 x t x t A d y t d y t z z t = + = + = + = + = = + c) 1 2 1 1 (1; 2;3), : 2 2 , : 2 3 3 3 x t x A d y t d y t z t z t = = = = + = = + d) 1 2 7 3 1 (4;1;4), : 4 2 , : 9 2 4 3 12 x t x t A d y t d y t z t z t = + = + = = + = + = HT 55. Vit phng trnh tham s ca ng thng i qua im A, vung gc v ct ng thng cho trc: a) (1;2; 2), : 1 2 x t A y t z t = = = b) 3 2 ( 4; 2;4), : 1 1 4 x t A d y t z t = + = = + c) 1 3 (2; 1; 3), : 1 2 2 x t A y t z t = + = + = + d) (3;1; 4), : 1 2 x t A y t z t = = = HT 56. Vit phng trnh tham s ca ng thng i qua im A v ct c hai ng thng d1, d2 cho trc: a) 1 2 1 2 1 (1;0;5), : 3 2 , : 2 1 1 3 x t x t A d y t d y t z t z t = + = = = + = + = b) 1 2 1 1 3 (2; 1;1), : 2 , : 2 3 3 x t x t A d y t d y t z z t = + = + = + = + = = + c) 1 2 1 3 2 2 ( 4; 5;3), : 3 2 , : 1 3 2 1 5 x t x t A d y t d y t z t z t = + = + = = + = = d) 1 2 1 3 (2;1; 1), : 2 4 , : 3 5 2 x t x t A d y t d y t z t z t = + = = + = = + = HT 57. Vit phng trnh tham s ca ng thng nm trong mt phng (P) v ct c hai ng thng d1, d2 cho trc: 251. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 19 a) 1 2 ( ) : 2 0 2 1 : , : 4 2 1 1 4 1 P y z x t x y z d d y t z + = = = = = + = b) 1 2 ( ) : 6 2 2 3 0 1 2 1 : 3 2 , : 2 1 1 3 P x y z x t x t d y t d y t z t z t + + + = = + = = = + = + = c) 1 2 ( ) : 2 3 3 4 0 7 3 1 : 4 2 , : 9 2 4 3 12 P x y z x t x t d y t d y t z t z t + = = + = + = = + = + = d) 1 2 ( ) : 3 3 4 7 0 1 1 : 2 2 , : 2 3 3 3 P x y z x t x d y t d y t z t z t + + = = = = = + = = + HT 58. Vit phng trnh tham s ca ng thng song song vi ng thng v ct c hai ng thng d1, d2 cho trc: a) 1 2 1 1 : 2 1 2 1 1 : 1 2 1 2 1 3 : 3 2 1 x y z x y z d x y z d = = + = = + + = = b) 1 2 1 5 : 3 1 1 1 2 2 : 1 4 3 4 7 : 5 9 1 x y z x y z d x y z d = = + = = + + = = c) 1 2 1 2 2 : 1 4 3 1 2 2 : 1 4 3 4 7 : 5 9 1 x y z x y z d x y z d + = = + = = + + = = d) 1 2 1 3 2 : 3 2 1 2 2 1 : 3 4 1 7 3 9 : 1 2 1 x y z x y z d x y z d + + = = + = = = = HT 59. Vit phng trnh tham s ca ng thng vung gc chung ca hai ng thng cho nhau d1, d2 cho trc: a) 1 2 3 2 2 3 : 1 4 , : 4 2 4 1 2 x t x t d y t d y t z t z t = = + = + = = + = b) 1 2 1 2 2 3 : 3 , : 1 2 2 3 4 4 x t x t d y t d y t z t z t = + = + = + = + = + = + c) 1 2 2 2 1 : 1 , : 3 3 1 2 x t x t d y t d y t z t z t = + = + = + = + = = + d) 1 2 2 3 1 2 : 3 , : 1 2 1 2 2 x t x t d y t d y t z t z t = + = + = = = + = + HT 60. Vit phng trnh tham s ca ng thng d l hnh chiu ca ng thng trn mt phng (P) cho trc: a) 2 3 1 : 2 1 3 ( ) : 2 2 3 0 x y z P x y z + = = + + = b) 3 2 2 : 1 2 3 ( ) : 3 4 2 3 0 x y z P x y z + = = + + = c) 1 1 3 : 1 2 2 ( ) : 2 2 3 0 x y z P x y z + = = + = d) 1 : 2 1 1 ( ) : 1 0 x y z P x y z = = + + = HT 61. Vit phng trnh tham s ca ng thng i qua im A, vung gc vi ng thng d1 v ct ng thng d2 cho trc: a) 1 2 1 1 2 (0;1;1), : , : 3 1 1 1 x x y z A d d y t z t = = = = = + 252. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 20 b) 1 2 2 1 1 (1;1;1), : , : 1 2 2 1 1 1 x x y z A d d y t z t = + = = = + = c) 1 2 1 4 1 1 3 ( 1;2; 3), : , : 6 2 3 3 2 5 x y z x y z A d d + + = = = = VN 2: V tr tng i gia hai ng thng xt VTT gia hai ng thng, ta c th s dng mt trong cc phng php sau: Phng php hnh hc: Da vo mi quan h gia cc VTCP v cc im thuc cc ng thng. Phng php i s: Da vo s nghim ca h phng trnh cc ng thng. BI TP C BN HT 62. Xt v tr tng i gia hai ng thng d1, d2 cho trc: a) {1 2 1 2 4 : ; : 1 ; ; 2 3 2 1 3 x y z d d x t y t z t + = = = + = = + b) { {1 2: 5 2 ; 1 ; 5 ; : 3 2 '; 3 '; 1 'd x t y t z t d x t y t z t= + = = = + = = c) { {1 2: 2 2 ; 1 ; 1; : 1; 1 ; 3d x t y t z d x y t z t= + = + = = = + = d) 1 2 1 2 3 7 6 5 : ; : 9 6 3 6 4 2 x y z x y z d d = = = = HT 63. Chng t rng cc cp ng thng sau y cho nhau. Vit phng trnh ng vung gc chung ca chng: a) { {1 2: 1 2 ; 3 ; 2 3 ; : 2 '; 1 '; 3 2 'd x t y t z t d x t y t z t= = + = = = + = b) { {1 2: 1 2 ; 2 2 ; ; : 2 '; 5 3 '; 4d x t y t z t d x t y t z= + = = = = = c) { {1 2: 3 2 ; 1 4 ; 4 2; : 2 3 '; 4 '; 1 2 'd x t y t z t d x t y t z t= = + = = + = = HT 64. Tm giao im ca hai ng thng d1 v d2: a) { {1 2: 3 ; 1 2 ; 3 ; : 1 '; 2 '; 4 'd x t y t z t d x t y t z t= = = + = + = = + b) {1 2 3 0 : ; : 1 ; 2 ; 3 2 1 0 x y z d d x t y t z t x y + + + = = + = + = + = HT 65. Tm m hai ng thng d1 v d2 ct nhau. Khi tm to giao im ca chng: a) { {1 2: 1 ; ; 1 2 ; : 1 '; 2 2 '; 3 'd x mt y t z t d x t y t z t= + = = + = = + = b) { {1 2: 1 ; 3 2 ; ; : 2 '; 1 '; 2 3 'd x t y t z m t d x t y t z t= = + = + = + = + = c) 1 2 2 4 0 2 3 0 : ; : 3 0 2 6 0 x y z x y mz d d x y x y z + = + + = + = + + = VN 3: V tr tng i gia ng thng v mt phng xt VTT gia ng thng v mt phng, ta c th s dng mt trong cc phng php sau: Phng php hnh hc: Da vo mi quan h gia VTCP ca ng thng v VTPT ca mt phng. 253. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 21 Phng php i s: Da vo s nghim ca h phng trnh ng thng v mt phng. BI TP C BN HT 66. Xt v tr tng i gia ng thng d v mt phng (P). Tm giao im (nu c) ca chng: a) {: 2 ; 1 ; 3 ; ( ) : 10 0d x t y t z t P x y z= = = + + + = b) {: 3 2; 1 4 ; 4 5; ( ) : 4 3 6 5 0d x t y t z t P x y z= = = = c) 12 9 1 : ; ( ) : 3 5 2 0 4 3 1 x y z d P x y z = = + = d) 11 3 : ; ( ) : 3 3 2 5 0 2 4 3 x y z d P x y z + = = + = e) 13 1 4 : ; ( ) : 2 4 1 0 8 2 3 x y z d P x y z = = + + = f) 3 5 7 16 0 : ; ( ) : 5 4 0 2 6 0 x y z d P x z x y z + + + = = + = g) 2 3 6 10 0 : ; ( ) : 4 17 0 5 0 x y z d P y z x y z + + = + + = + + + = HT 67. Cho ng thng d v mt phng (P). Tm m, n : i) d ct (P). ii) d // (P). iii) d (P). iv) d (P). a) 1 2 3 : ; ( ) : 3 2 5 0 2 1 2 x y z d P x y z m m + + = = + = b) 1 3 1 : ; ( ) : 3 2 5 0 2 2 x y z d P x y z m m + = = + + = HT 68. Cho ng thng d v mt phng (P). Tm m, n : a) {: ; 2 ; 3d x m t y t z t= + = = ct ( ) : 2 5 0P x y z + = ti im c tung bng 3. b) 2 3 0 : 2 5 0 x y d y z = + + = ct ( ) : 2 2 2 0P x y z m+ + = ti im c cao bng 1. c) 2 3 0 : 3 2 7 0 x y d x z + = = ct ( ) : 0P x y z m+ + + = VN 5: Khong cch 1. Khong cch t im M n ng thng d Cch 1: Cho ng thng d i qua M0 v c VTCP a . 0 , ( , ) M M a d M d a = Cch 2: Tm hnh chiu vung gc H ca M trn ng thng d. d(M,d) = MH. Cch 3: Gi N(x; y; z) d. Tnh MN2 theo t (t tham s trong phng trnh ng thng d). 254. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 22 Tm t MN2 nh nht. Khi N H. Do d(M,d) = MH. 2. Khong cch gia hai ng thng cho nhau Cho hai ng thng cho nhau d1 v d2. d1 i qua im M1 v c VTCP 1a , d2 i qua im M2 v c VTCP 2a 1 2 1 2 1 2 1 2 , . ( , ) , a a M M d d d a a = Ch : Khong cch gia hai ng thng cho nhau d1, d2 bng khong cch gia d1 vi mt phng () cha d2 v song song vi d1. 3. Khong cch gia hai ng thng song song bng khong cch t mt im thuc ng thng ny n ng thng kia. 4. Khong cch gia mt ng thng v mt mt phng song song Khong cch gia ng thng d vi mt phng () song song vi n bng khong cch t mt im M bt k trn d n mt phng (). BI TP C BN HT 69. Tnh khong cch t im A n ng thng d: a) 1 4 (2;3;1), : 2 2 4 1 x t A d y t z t = = + = b) 2 2 (1;2; 6), : 1 3 x t A d y t z t = + = = c) 2 1 (1;0;0), : 1 2 1 x y z A d = = d) 2 1 1 (2;3;1), : 1 2 2 x y z A d + + = = HT 70. Chng minh hai ng thng d1, d2 cho nhau. Tnh khong cch gia chng: a) { {1 2: 1 2 ; 3 ; 2 3 ; : 2 '; 1 '; 3 2 'd x t y t z t d x t y t z t= = + = = = + = b) { {1 2: 1 2 ; 2 2 ; ; : 2 '; 5 3 '; 4d x t y t z t d x t y t z= + = = = = = c) { {1 2: 3 2 ; 1 4 ; 4 2; : 2 3 '; 4 '; 1 2 'd x t y t z t d x t y t z t= = + = = + = = HT 71. Chng minh hai ng thng d1, d2 song song vi nhau. Tnh khong cch gia chng: a) { {d1 2: 3 2 , 4 3 , 2 ; : 4 4 , 5 6 , 3 2d x t y t z t x t y t z t= + = + = + = + = + = + b) 1 2 1 2 3 2 3 1 : ; : 2 6 8 3 9 12 x y z x y z d d + + + = = = = HT 72. Chng minh ng thng d song song vi mt phng (P). Tnh khong cch gia chng: a) {: 3 2; 1 4 ; 4 5; ( ) : 4 3 6 5 0d x t y t z t P x y z= = = = b) {: 1 2 ; ; 2 2 ; ( ) : 8 0d x t y t z t P x z= = = + + + = c) 2 1 0 : ; ( ) : 2 2 4 5 0 2 3 0 x y z d P x y z x y z + + = + + = + = 255. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 23 VN 6: Gc 1. Gc gia hai ng thng Cho hai ng thng d1, d2 ln lt c cc VTCP 1 2,a a . Gc gia d1, d2 bng hoc b vi gc gia 1 2,a a . ( ) 1 2 1 2 1 2 . cos , . a a a a a a = 2. Gc gia mt ng thng v mt mt phng Cho ng thng d c VTCP 1 2 3( ; ; )a a a a= v mt phng () c VTPT ( ; ; )n A B C= . Gc gia ng thng d v mt phng () bng gc gia ng thng d vi hnh chiu d ca n trn (). ( ) 1 2 3 2 2 2 2 2 2 1 2 3 sin ,( ) . Aa Ba Ca d A B C a a a + + = + + + + BI TP C BN HT 73. Tnh gc gia hai ng thng: a) { {1 2: 1 2 , 1 , 3 4 ; : 2 , 1 3 , 4 2d x t y t z t d x t y t z t= + = + = + = = + = + b) 1 2 1 2 4 2 3 4 : ; : 2 1 2 3 6 2 x y z x y z d d + + + = = = = c) {1 2 2 3 3 9 0 : ; : 9 ; 5 ; 3 2 3 0 x y z d d x t y t z t x y z = = = = + + + = HT 74. Chng minh hai ng thng sau vung gc vi nhau: a) 1 2 7 2 15 0 7 0 : ; : 7 5 34 0 3 4 11 0 x z x y z d d y z x y = = + + = = HT 75. Tm m gc gia hai ng thng sau bng : a) { { 0 1 2: 1 ; 2; 2 ; : 2 ; 1 2; 2 ; 60d x t y t z t d x t y t z mt = + = = + = + = + = + = . HT 76. Tnh gc gia ng thng d v mt phng (P):: a) 1 1 3 : ; ( ) : 2 2 10 0 1 2 3 x y z d P x y z + = = = . b) { 4 4 : 1; 2 5; 3 ; ( ) : 5 4 0d x y t z t P x z= = + = + + + = c) 4 2 7 0 : ; ( ) : 3 1 0 3 7 2 0 x y z d P x y z x y z + + = + + = + = VN 7: Mt s vn khc 1. Vit phng trnh mt phng Dng 1: Mt phng (P) i qua im A v ng thng d: Trn ng thng d ly hai im B, C. 256. GV.Lu Huy Thng 0968.393.899 B HC V B - CHUYN CN S N BN Page 24 Mt VTPT ca (P) l: ,n AB AC = . Dng 2: Mt phng (P) cha hai ng thng song song d1, d2: Xc nh VTCP a ca d1 (hoc d2). Trn d1 ly im A, trn d2 ly im B. Suy ra A, B (P). Mt VTPT ca (P) l: ,n a AB = . Dng 3: Mt phng (P) cha hai ng thng ct nhau d1, d2: Ly im A d1 (hoc A d2) A (P). Xc nh VTCP a ca d1, b ca d2. M