Cac Bai Toan Boi Duong Hsg 2010

8/8/2019 Cac Bai Toan Boi Duong Hsg 2010 WWW.VNMATH.COM

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S GD&T NGH AN TRNG THPT NG THC HA

MT S BITON CHN LC BI DNGHC SINH GIIMN TON

VIT BI: PHM KIM CHUNG THNG 12 NM 2010

PHN MC LC Trang

I PHNG TRN H BPT HPT CC BI TON LIN QUAN N O HM

II PHNG TRN H HM VA THC

III BT NG THC V CC TR

IV GII HN CA DY S

V HNH HC KHNG GIAN

VI T LUYN V LI GII

DANH MC CC TI LIU THAM KHO

1. Cc din n :www.dangthuchua.com ,www.math.vn,www.mathscope.org,www.maths.vn,www.laisac.page.tl , www.diendantoanhoc.net,www.k2pi.violet.vn ,www.nguyentatthu.violet.vn ,

2. thi HSG Quc Gia, thi HSG cc Tnh Thnh Ph trong nc, thi Olympic 30 -43. B sch : Mt s chuyn bi dng hc sinh gii ( Nguyn Vn Mu Nguyn Vn Tin ) 4. Tp ch Ton Hc v Tui Tr 5. B sch : CC PHNG PHP GII (Trn Phng - L Hng c )6. B sch : 10.000 BI TON S CP (Phan Huy Khi )7. B sch : Ton nng cao ( Phan Huy Khi )8. Gii TON HNH HC 11 ( Trn Thnh Minh )9. Sng to Bt ng thc ( Phm Kim Hng )10. Bt ng thc Suy lun v khm ph ( Phm Vn Thun ) 11. Nhng vin kim cng trong Bt ng thc Ton hc ( Trn Phng ) 12. 340 bi ton hnh hc khng gian ( I.F . Sharygin ) 13. Tuyn tp 200 Bi thi V ch Ton ( o Tam ) 14. v mt s ti liu tham kho khc . 15. Ch : Nhng dng ch mu xanhcha cc ng link n cc chuyn mc h oc cc website.

httP://www.vnmath.com Dich vu Toan ho

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2/46

PHN I : PHNG TRNH BPT - H PTV CC BI TON LIN QUAN N O HM1. = + + +2y 2x 2 m 4xx 5Tm cc gi tr ca tham s m hm s : c cc i . S : m < -22. + =/=

=

3 21 x sin 1, xf(x)

0 , x 0

x 0Cho hm s : . Tnh o hm ca hm s ti x = 0 v chng minh hm s t cc tiu

ti x =0 .

3. ( )= = y f(x) | x | x 3Tm cc tr ca hm s : . S : x =0 ; x=14. Xc nh cc gi tr ca tham s m cc phng trnh sau c nghim thc :

( ) ( )+ + + =x 3 3m 4 1 x3 m4 1m 0 a) . S : 7

9

9m

7

+ =4 2x 1 x mb) . S :


3/46

22.Gii h PT :( ) ( )

=

=

4 4

3 3 2 2

x y 240

x 2y 3 x 4y 4 x 8y

23.Gii h phng trnh :( )

+ + = + +

=

4 3 3 2 2

3 3

x x y 9y y x y x 9x

x y x 7 . S : (x,y)=(1;2)

24. Gii h phng trnh : ( ) ( ) + + = + + =

2

2 2

4x 1 x y 3 5 2y 0

4x y 2 3 4x 7

25.Tm m h phng trnh sau c nghim : + + = + =

2 xy y x y 5

5 x 1 y m. S : m 1; 5

26.Xc nh m phng trnh sau c nghim thc : ( ) ( ) + + + =

41

x x 1 m x x x 1 1x 1

.

27.Tm m h phng trnh : ( ) + + =+ =

2

3 x 1 y m 0

x xy 1

c ba cp nghim phn bit .

28.Gii h PT :

+ + = +

+ + = +

2 y 1

2 x 1

x x 2x 2 3 1

y y 2y 2 3 1

29. ( thi HSG Tnh Ngh An nm 2008 ) .Gii h phng trnh : =

= +

x y sinxesiny

sin2x cos2y sinx cosy 1

x,y 0;4

30. Gii phng trnh : + =3 2 316x 24x 12x 3 x 31.Gii h phng trnh : ( )

( )

+ + + = +

+ + + + =

2x y y 2x 1 2x y 1

3 2

1 4 .5 2 1

y 4x ln y 2x 1 0

32.Gii phng trnh : ( )= + + +x 33 1 x log 1 2x 33. Gii phng trnh : + + =

33 2 2 3

2x 10x 17x 8 2x 5x x S

34. Gii h phng trnh : + = ++ + + =

5 4 10 6

2

x xy y y

4x 5 y 8 6

35. Gii h phng trnh : + + = + ++ + = + +

2 2

2 2

x 2x 22 y y 2y 1

y 2y 22 x x 2x 1

36. Gii h phng trnh :

+ =

+ = +

y x

1x y

2

1 1x y

y x

37. ( thi HSG Tnh Qung Ninh nm 2010 ) . Gii phng trnh : =

2 21 1x5x 7

( x 6)x

51

Li gii : K : >7

x5

Cch 1 : PT

+ = = +

4x 6 36(4x 6)(x 1) 0 x

2(x 1)(5x 7). x 1 5x 7

Cch 2 : Vit li phng trnh di dng : ( ) =

2 21 15x 6 x(5x 6) 1 x 1

V xt hm s : = >

21 5

f(t) t , t 7t 1


3
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4/46

38.( thi HSG Tnh Qung Ninh nm 2010 ) Xc nh tt c cc gi tr ca tham s m BPT sau c nghim :+ 3 2 33x 1 m( x x 1)x

HD : Nhn lin hp a v dng : ( )+ + 3

3 2x x 1 (x 3x 1) m

39.( thi HSG Tnh Qung Bnh nm 2010 ) . Gii phng trnh :+ + + = + +3 2x 3x 4x 2 (3 2) 3xx 1

HD : PT ( ) + + ++ = + +3

3(x 1) (x 1) 3x 1 3x 1 . Xt hm s : = + >3 tf t) t ,t( 0

40.

( thi HSG Tnh Hi Phng nm 2010 ) . Gii phng trnh : = + 3 23 2x 1 27x 27x 13x2 2

HD : PT = + + = 33 32x 1 (3x 1) 2(2x 1) 2 (3x 1) f( 2x 1) f(3x 1)

41.( thi Khi A nm 2010 )Gii h phng trnh : + + =+ + =

2

2 2

(4x 1)x (y 3) 5 2y 0

4x y 2 3 4x 7

HD : T pt (1) cho ta : ( ) + + = = 2

2 1].2x 5 2y 5 2y f([(2x 2x) f(1 5) 2y )

Hm s : + == + > 2 21).t f '( t) 3t f(t) (t 1 0

= = =225 4x

2x 5 2y 4x 5 2y y2

Th vo (2) ta c :

+ + =

22

2 5 4x

4x 2 3 4x 72 , vi 0

3

x 4 ( Hm ny nghch bin trn khong ) v c

nghim duy nht : =x1

2.

42.( thi HSG Tnh Ngh An nm 2008 ) . Cho h: + =+ + +

x y 4

x 7 y 7 a(a l tham s).

Tm a h c nghim (x;y) tha mn iu kin x 9. HD : ng trc bi ton cha tham s cn lu iu kin cht ca bin khi mun quy v 1 bin kho st :

= x y 0 x4 16 . t = x , t [t 3;4] v kho st tm Min .S : +a 4 2 2

43. Gii h phng trnh : + + =+ = +

4 xy 2x 4

x 3 3 y

y 4x 2 5

2 x y 2

44. Xc nh m bt phng trnh sau nghim ng vi mi x : ( ) + 2sinx sinx sinxe 1 (e 1)sinx2e e 1e 1 45.( thi HSG Tnh Tha Thin Hu nm 2003 ) . Gii PT :

+ + = 2 2

2 5 2 2 5log (x 2x 11) log (x 2x 12)

46.nh gi tr ca m phng trnh sau c nghim: ( ) ( ) + + + =4m 3 x 3 3m 4 1 x m 1 0

47.(Olympic 30-4 ln th VIII ) . Gii h phng trnh sau: +

=+

+ + = + + +

2 22

y x

2

3 2

x 1e

y 1

3log (x 2y 6) 2log (x y 2) 1

48.Cc bi ton lin quan n nh ngh a o hm : Cho +

>=

+

x

2

(x 1)e , x 0f(x)

x ax 1, x 0

. Tm a tn ti f(0) .

Cho += + + = =

2 2x x

lnx , x 0F(x) 2 4

0, , x 0

v>

= =

xlnx , x 0f(x)

0, x 0. CMR : =F'(x) f(x)

Cho f(x) xc nh trn R tha mn iu kin : >a 0bt ng thc sau lun ng x R : + < 2| f(x a) f(x) a | a. Chng minh f(x) l hm hng .


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mailto:[email protected]


5/46

Tnh gii hn :

=

x

3

1 2

4

tanN lim

2sin

x 1

x 1 Tnh gii hn :

+=

+

2 32x 2

2 2x 0

e 1N lim

ln(1 x

x

)

Tnh gii hn :

+ + =

+ 33

x 0

3 32x x 1N

1m

xli

x Tnh gii hn :

=

sin2x

4

s

x

nx

0

ie eN lim

sinx

Tnh gii hn :

+=

0

3

5x

x 8 2

siN lim

n10x Tnh gii hn :

+=

+

2 32x 2

6 2x 0

e 1N lim

ln(1 x

x

)

Tnh gii hn : =sin2x sin3

7x

3x

0

eN lim

e

sin4x Tnh gii hn : =x 4

3x 0 38

4 x

N xim 2l

Tnh gii hn :

=

+ 9

x 0

3x 2x.3 cos4x

1 sinx 1

2N lim

sinx

Cho P(x) l a thc bc n c n nghim phn bit1 2 3 n

x x x; ; ...x . Chng minh cc ng thc sau :

a) + + + =2 n2 n

1

1

P''(x ) P''(x ) P''(x )... 0

P'(x P'( P'(x) )x)

b) + + + =2 n1 ) )

1 1 1... 0

P'(x P'(x P'(x )

Tnh cc tng sau :a) = + + +

nT osx 2cos2x ... nc(x) c osnx

b) = + + +n 2 2 n n

1 x 1 x 1 x(x) tan tan ... tan

2 2 2 2 2 2T

c) + + + = 2 3 n n 2n n n

CMR : 2.1.C 3.2.C ... n(n 1)C n(n 1).2

d) + + + += 2n

S inx 4sin2x 9sin3x ...(x) s sn innx

e) + + + = + + ++ + + + +

n 2 2 2 2 2 2

2x 1 2x 3 2x (2n 1)(x) ...

x (x 1) (x 1) (x 2) x (n 1) (x n)S

49. Cc bi ton lin quan n cc tr ca hm s : a) Cho + R: a b 0 . Chng minh rng :

+ +

n na b a b

2 2

b) Chng minh rng vi >a 3, n 2 ( n N, nchn ) th phng trnh sau v nghim :+ + ++ + + =n 2 n 1 n 2(n 1)x 3(n 2)x a 0

c) Tm tham s m hm s sau c duy nht mt cc tr : + +

= + +

22 2

2 2y (m 1) 3

x x

1 x 1 xm 4m

d) Cho n 3,n N ( n l ) . CMR : =/x 0 , ta c : + + + + +


6/46

PHN II : PHNG TRNH HM-A THC

1. Tm hm s : f : R R tho mn ng thi cc iu kin sau :a)

=

x 0

f(x)lim 1x

b) ( ) ( ) ( )+ = + + + + 2 2f x y f x f y 2x 3xy 2y , x,y R 2. Tm hm s : f : R R tho mn iu kin sau : ( ) ( ) ( ) = + + + + 2008 2008f x f(y) f x y f f(y) y 1, x,y R 3. Tm hm s : f : R R tho mn iu kin sau : ( ) ( ) ( )( )+ = + f x cos(2009y ) f x 2009cos f y , x ,y R 4. Tm hm s : f : R R tho mn ng thi cc iu kin sau :

c) ( ) 2009xf x e d) ( ) ( ) ( )+ f x y f x .f y , x,y R

5. Tm hm s : f : R R tho mn iu kin sau : ( ) ( )+ = f y 1f x y f(x).e , x,y R 6. Tm hm s : f : R R tho mn iu kin sau : ( )( ) ( )+ = + 2f x.f x y f(y.f x ) x 7. ( thi HSG Tnh Hi Phng nm 2010 ) Tm hm f : tha mn :

( )+ + = + 2(x) 2yf(x) f(y) f y f(x) , ,x,yf R


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7/46

PHN III : BT NG THCV CC TR

1. Cho + + =2 2 2a,b,c R: a b c 3 . Chng minh rng : + + 2 2 2a b b c c a 3 2. Cho cc s thc khng m a,b,c . Chng minh rng :

( ) ( ) ( ) ( ) ( ) ( ) + + 2 2 2 2 2 22 2 2 2 2 2

a b a b b c b c c a c a a b b c c a

3.Cho cc s thc a,b,c. Chng minh rng : ( ) ( )+ + + + ++

2 2 2 2

2

a b c 81 a b 13

a b cb c a 4 42a b

4. Cho cc s thc khng m a,b,c tho mn : + + + =a b c 36abc 2 . Tm Max ca : = 7 8 9P a b c 5. Cho 3 s thc dng tu x,y,z . CMR : + +

+ + +a b c 3

a b b c c a 2

6. Cho a,b,c >0 . Tm GTNN ca : ( )+ +=6

2 3

a b cP

ab c

7. Cho cc s thc dng x,y,z tha mn : + + =2 2 2yx z 1 CMR:

+ +

2 2 22x (y z) 2y (z x) 2z (x y)

yz zx xy

8. Cho cc s thc dng a,b,c .CMR : + ++ + + + + + + +bc ca ab a b ca 3b 2c b 3c 2a c 3a 2b 6 9. Cho cc s thc dng a,b,c . CMR : + +

+ + + + + +3 3 3 3 3 31 1 1 1

abca b abc b c abc c a abc

10. Cho cc s thc tha mn iu kin : + + =+ + +2 2 2

1 1 11

a 2 b 2 c 2. CMR : + + ab bc ca 3

11. Cho cc s thc dng tha mn iu kin : + + =2 2 2ba c 3 . CMR :+ +

1 1 13

2 a 2 b 2 c

12. Cho x,y,z l 3 s thc dng ty . CMR : + + + + +x y z 3 2

x y y z z x 2

13. Cho cc s thc dng a,b,c . CMR: + + + + ++ +

2 2 2 2a b c 4(a b)a b c

b c a a b c

14. Cho cc s thc dng a,b,c tha mn : abc=1 . CMR : + + + + +3 3 3

1 1 1 3

2a (b c) b (c a) c (a b)

15. Cho 3 s thc x,y,z tha mn : xyz=1 v ( )( )( ) = /x 1 y 1 z 1 0 . CMR :

+ +

22 2x y z

1x 1 y 1 z 1

16. Cho a,b,c l cc s thc dng bt k .CMR: + + ++ + + + + + + +

2 2 2

2 2 2 2 2 2

(3a b c) (3b c a) (3c a b) 9

22a (b c) 2b (c a) 2c (a b)

17. Cho cc s thc dng a,b,c tha mn : + + =2 2 2ba c 1 . CMR :+ +

1 1 1 9

1 ab 1 bc 1 ca 2

18. Cho cc s thc a,b,c tha mn : + + =2 2 2ba c 9 .CMR : ++ +2(a b c) 10 abc 19. Cho a,b,c l cc s thc dng : a+b+c =1 . CMR : + +

3 3 3

2 2 2

a b c 1

4(1 a) (1 b) (1 c)

20.(Chn THSG QG Ngh An nm 2010 ) Cho cc s thc dng a,b,c tha mn :+ + + + + =4 4 4 2 2 2b c ) 25(9(a a b c ) 48 0 . Tm gi tr nh nht ca biu thc :

+ +=+ + +

2 2 2a b c

b 2c c 2a aF

2b


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8/46

Li gii :

T gi thit :

+ + + + + = + +

= + + + + + +

+ + + + + + +

4 4 4 2 2 2 2 2 2 4 4 4 2 2 2 2

2 2 2 2 2 2 2 2 2 2

b c ) 25(a b c ) 48 0 25(a b c ) 48 9(a b c ) 48 3(a b c )

3(a b c ) b c ) 48 0

9

3 b c

(a

1625(a a

3

Ta li c :

+ ++ + = + +

+ + + + + + + + + + +=

4 4 42 2 2 2 2 2 2

2 2 2 2 2 2 2 2 2

a b c a b c (a b c )

b 2c c 2a a 2b a (b 2c) b (c 2a) c (a 2b) (a b b c c a) 2(a c b a cF

b)

Lic : + ++ + = + + + + + + + +2 2 2 2

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 (a b c )b b c c a a(ab) b(bc) c(ca) (a b c ) b c ca [a b a ] a b c3

Tng t :+ +

+ + + +2 2 2

2 2 2 2 2 2 a b cc b a c b) a b c .(a3

T ta c :+ +

2 2 2

Fa b c

13

. Du bng xy ra khi v ch khi : a=b=c=1.

P N CA S GD&T NGH AN

p dng bt ng thc AM GM, ta c

+ ++ =

+ +

2 2 2 2 2a (b 2c)a a (b 2c)a 2a2

b 2c 9 b 2c 9 3.

Tng t+ +

+ + + +

2 2 2 2 2 2b (c 2a)b 2b c (a 2b)c 2c,

c 2a 9 3 a 2b 9 3.

Suy ra: = + ++ + +

2 2 2a b c

Fb 2c c 2a a 2b

( ) + + + + + + + 2 2 2 2 2 22 1a b c a (b 2c) b (c 2a) c (a 2b) (*)

3 9.

Li p dng AM GM, ta c

+ + + + + ++ + + + = + +

3 3 3 3 3 3 3 3 32 2 2 3 3 3a a c b b a c c b

a c b a c b a b c (**)3 3 3

.

T (*) v (**) suy ra:

( ) ( ) + + + + + +2 2 2 2 2 22 1F a b c a b c (a b c )3 9 ( ) ( ) ( ) + + + + + +2 2 2 2 2 2 2 2 22 1a b c a b c 3 a b c

3 9 .

t ( )= + +2 2 2t 3 a b c , t gi thit ta c:

( ) ( ) ( )+ + = + + + +2

2 2 2 4 4 4 2 2 225 a b c 48 9 a b c 3 a b c

( ) ( ) + + + + + + + 2

2 2 2 2 2 2 2 2 2 163 a b c 25 a b c 48 0 3 a b c3

.

Do =2 32 1

F t t f(t)9 27

vi t 3; 4 (* * *) .

M

= =t 3;4min f(t) f(3) 1 (** **) . T (***) v (****) suy ra F 1.

Vy =minF 1 xy ra khi = = =a b c 1 .

21. ( thi HSG Tnh Ngh An nm 2009 ) Cho cc s thc dng x,y,z . Chng minh rng : + +

+ + +2 2 2 2 2 21 1 1 36

x y z 9 x y y z z x

Li gii :

BT cho tng ng vi : ( )

+ + + + +

2 2 2 2 2 2 1 1 19 x y y z z x 36x y z

Ta c : ( )+ +

=

32 xy yz zx

xyz (xy)(yz)(zx)3


8


9/46

Do :( )+ + + +

+ + = = + ++ +

22 2

3

27 xy yz zx1 1 1 xy yz zx 27

x y z xyz xy yz zx(xy yz zx)

Li c : ( )+ + + + + + + + ++ = + +2 2 2 2 2 2 2 2 2 2 2 2y y z z x y 1 z 1) (z x 1) 29 x 6 x (y 3 (xy yz zx)

Nn :

( )

+ + + = + + + + + + + +

22 27 9VT 4 3 (xy yz zx) . 108 6 (xy yz zx)

xy yz zx xy yz zx

+ + + = + +

9

108 6 2 (xy yz zx) 1296 VT 36xy yz zx

P N CA S GD&T NGH AN :

Bt ng thc cn chng minh tng ng(xy + yz + zx)(9 + x2y2+ z2y2+x2z2) 36xyz

p dng bt ng thc Csi ta c :

xy + yz + zx 3 2 2 23 x y z (1)

V 9+ x2y2+ z2y2+x2z2 4 4 412 x y z 12 hay 9 + x2y2+ z2y2+x2z2 3 xyz 12 (2)

Do cc v u dng, t (1), (2) suy ra:

(xy + yz + zx)(9 + x2y2+ z2y2+x2z2) 36xyz (pcm).Du ng thc xy ra khi v ch khi x = y = z =1

22.( thi HSG Tnh Qung Ninh nm 2010 ) Cho cc s thc dng x,y tha mn k : + + =x y 1 3xy . Tm gi trln nht ca : = +

+ + 2 23x 3y 1

My(x 1) x y 1) x

1

y(

Li gii :

Ta c : = + + + 3xy x y 1 2 xy 1 xy 1 xy 1 (*)Ta c :

( )+= +

+ +

+ += + = =

22

2 2 2 2 2 2 2 2 22

3xy 3xy 1 (1 3xy )1 1 1 3xy(x y) (x y)

y y (3

3x 3y 1 2xyM

y (3x 1) x (3y 1) x 9xy 3x 1) x (x y(3y 1) x y 4x) y123.( thi HS G Tnh Qung Bnh nm 2010 ) Cho cc s thc dng a, b, c . CMR :

+ + ++3 3

3 3

3

3

c a b c

b c aa

a b

b c

HD :

+ +

+ +

3 3

3 3

3 3 3

3 3 3

a a1

b b

a b c3

b c a

a3

b

24.( thi HSG Tnh Vnh Phc nm 2010 ) . Cho x, y, z 0 tha mn : + + =2 2 2yx z 1 . Tm gi tr ln nht cabiu thc : = + +P 6(y z x ) 27xyz

HD :+ + + = +

2 2 22 2 2y z 1 x6 2(y z ) x 27x. 6 2(1 x ) x 27x

2P

2 ( )=MaxP 10

25.( thi HSG Tnh Hi Phng nm 2010 ) . Cho + + =2 2 20: a bb,c ca, 1 . Chng minh rng :+ + 3 3 3

62b 3ca

7

HD : C thdng cn bng h s hoc Svacx 26. Cho x,y,z l cc s thc dng tha mn : =xyz 1 . Chng minh rng :

+ ++ + +

+ + +

4 4 3 4 4 3 4 4 3

6 6 6 6 6 6

(x (y (z

x y

y ) z ) x )12

y xzz


9


10/46

Li gii : t = = = =2 2 2a;y b;z cx abc 1 . Bt ng thc cho tr thnh :

+ ++ + +

+ + +

3 3 3

3 3

2 2 2 2 2

3 3 3

2

3

(a (b (c

a b

b ) c ) a )12

b acc

p dng Bt ng thc AM-GM cho 4 s ta c :

( ) ( ) ( )= + + + ++ + + ++ 42 2 3 6 4 2 4 2 4 2 6 2 4 2 4 2 4 6 6 3 3(a ab ) b a b a b b b b a b ab a a a 4 ba 27. ( thi HSG Tnh ng Nai nm 2010 ). Cho a,b,c > 0 . Chng minh rng :

+ ++ +

+

+ + ++

2 2 2

1 1 1 3(a b c)

a b b c c a b2( ca )

HD :

BT+ + + + + + + + + + +

+2 2 2 2 2 2(a 1b ) (b c ) 1 1 3(a b c)2 a

(c a

b b c a

)

c 2

V ch :+

+ 2

2 2 (a b)a b2

28.( thi HSG Tnh Ph Th nm 2010 ) . Cho > + + =x,y,z 0 : x y z 9 . Chng minh rng : + +

+ ++ + +

+

3 3 3 3 3 3x y z

xy 9 yz 9 zx

z x

9

y9

29.( thi chn T Ninh Bnh nm 2010 ) . Cho a,b,c l di ba cnh mttam gic c chu vi bng 4. Chng minhrng : + + +

2 2 2 272

a 2abcb c 27 HD : Bi ny th chn phn t ln nht m o hm .

30.( thi HSG Tnh Bnh nh nm 2010 ) . Cho a,b,c >0 . CMR : + + + +3 3 3b c aaca abc

bb

c

HD : + + + + + += 4 2 2 2 2 4a (a b c ) (a b c)

a b cabc 3abc 27abc

VT

31. ( thi chn HSG QG Tnh Bnh nh nm 2010) . Cho x,y,z >0 tha mn : + =2 xy xz 1 . Tm gi tr nhnht ca : += +

3yz 4zS

x 5xy

x y z

32.( thi chn HSG Thi Nguyn nm 2010 ). Cho cc s thc x,y,z tha mn iu kin : + + =+ + +1 2 3

11 x 2 y 3 z

.

Tm gi tr nh nht ca : =P xyz

33.( thi chn HSG QG tnh Bn Tre nm 2010 ) . Cho > + + =2 2 2ba,b c :a c, 0 3 . Chng minh bt ng thc :+ +

1 1 11

4 ab 4 bc 4 ca

34.( thi chn T trng HSP I H Ni 2010). Cho cc s thc dng x,y,z . Tm gi tr nh nht ca :+ +

+ += +

2 2 2

3 3 3 2 2 2

y y z z x 1xP

3xyz

x y 3(xy yz zxz )

Li gii 1 :

t : = = = =x y z

a; b; c abc 1y z x

. Lc : += ++ +

+2 2 2

b c 13

3

aP

b (a b c)c a

Ta c : + ++ + = + + = + + 2(ab bc ca)(a b c) abc(a b c) (ab)(ac) (ab)(bc) (ac)(bc)

3

Li c :

+ + + = + +

+

+ +

+

2

2 2 2 2

2

2b

b

1 a 1

a b

1 b 1 a

b

c 1 1 12 ab bc ca

a b cc acc b

1 c 12

c ca

Do : + + ++ + 2

13(ab bc ca)

(ab bc ca)P ( Vi + + ab bc ca 1 )


10


11/46

Li gii 2 :

t : = = = =z

a; b; c ay

z

x

ybc

x1 . Lc : + + + + +

+ ++

+=

+

2 2 2

2

b c 13abc 13(a b c)

c a 3(ab b

aP

c ca) (a b c)b

35.Bi ton tng t : Cho >x,y ,z 0: xyz 1 . Chng minh rng : + + + + +2 2 2

x y z 34

x y zy z x

Li gii : t : = = = 1 1 1

a; b; c abc 1x y z

.

BT cho tr thnh :+ +

+ + + ++ + + + + +

2 2 2 2

2

a b c 3abc (a b c) 9

c a b ab bc ca a b c (a b c) . Vi : + =+3

3a abcb c 3

36.( thi chn i tuyn H Vinh nm 2010 ) . Cho a,b,c l cc s thc thuc on [0;1]v + + =a b c 1 . Tm gitr ln nht v nh nht ca : = +

++

+ +2 2 21 1 1

Pa b c1 1 1

HD : Dng pp tip tuyn v Bt ng thc : + + + +

++ +2 2 2

1 1 11 ,

x y (x yx,y 0;

)x y 1

1 1 1

37.( thi chn HSG Q G tnh Lm ng ) . Cho a,b,c l cc s thc dng . Chng minh rng :+ + + + + ++

2 2 22 2 2 2 2 2b c a ab b b bc c c ca

c aba

a

Li gii :

C1 : ( THTT) Ta c : + + + + + + + +

+

+ +

2 2 2 2 2 2

b c b cc a 2(a b c) a b cc b a

a bb a c

a

Do :

= + + + = +

++

2 2 2 2 2 2b c a a2.VT 2 b a b b 2VP

c a

a

b b

ab b

b

C2 : Ta c : + + + 2 2a ab b a b c(Mincopxki)

M : + +

+= + +

Sv

2 22 22

acx

2

o

aaVT a

b

ab bab bab

ab

b c

38.( thi chn i tuyn trng Lng Th Vinh ng Nai nm 2010 ) . Cho > =a,b,c 0 : abc 1 . Chng minhrng : ++ + +2 2 2ab bc c a ba c

HD : BT + + + +

a b c

a b cb c a . Ch l :

=+ + 2 2 a

c 3a a c b

a b

ab c

Li gii 2 : Ta c : + + =2 2 2 2 2 2 33ab 3 (a )bab bc b c 3b

39.( Chn T HSG QG tnh Ph Th nm 2010 ). Cho >a,b,c 0 . Chng minh bt ng thc : + + +

+ +

3

3 3

2

3

2 2a b c

b c c a b

3 2

a 2

HD : + + + + + + + +

2 2

3 3

3

b c b c b c a 1 a2 3 2

a a a 2(a b c) b c3 2

40.( thi HSG Tnh Ngh An nm 2008 ) . Cho 3 s dng a,b,c thay i . Tm gi tr ln nht ca : = + +

+ + +

bc ca abP .

a 3 bc b 3 ca c 3 ab

HD : t = = = =b c

x; y; zc a

axyz

b1 . Lc :

= + + = + + + + +

z x y 1 x

P 1x 3z y 3x z 3y 3 x 3z

. Li c :

+ + + += =

+ + + + + + + + ++ + +

2 2 2

2 2 22

x x (x y z) (x y z) 3

x 3z 4x 3zx (x y z) (xy yz zx) (x y z)(x y z)

3

Do : =1 3 3

P 13 4 4

. Du = xy ra khi v ch khi : x = y = z =1 .


11


12/46

P N CA S GD&T :

t ( )= = = + x a ,y b ,z c;x,y,z 0; .

Khi : = + ++ + +2 2 2yz zx xy

P .x 3yz y 3zx z 3xy

Ta c = + ++ + +2 2 2

3yz 3zx 3xy3P

x 3yz y 3zx z 3xy

= + + =

+ + +

2 2 2

2 2 2

x y z3 3 Q

x 3yz y 3zx z 3xy

p dng bt BCS ta c

( )

+ + + + + + + +

+ + + + +

2

2 2 2

2 2 2

2 2 2

x y zx 3yz y 3zx z 3xy

x 3yz y 3zx z 3xy

Q. x y z 3xy 3yz 3zx

( )

( )

+ +

+ + + + +

2

2

x y zQ

x y z xy yz zx. Mt khc

( )+ ++ +

2x y z

xy yz zx3

Suy ra 3Q4

,do 9 33P P .4 4

Du bng xy ra khi v ch khi = =a b c. Vy gi tr nh nht ca P bng3

.4

41.( d b HSG Tnh Ngh An 2008 ) . Cho ba s dng a,b,c tho mn : + + =2 2 2a b c 1 . Tm gi tr nh nhtca biu thc : = + +

+ + +

2 2 2a b c

P .b c c a a b

Li gii 1: Gi s : + + +

1 1 1b c

b c c a aa

b. p dng bt ng thc Chebysev ta c :

( )

= + + + + = + + + + + + + + + + + + +

+ +

+ +

2

2

2 2

2 2

2

2

2a b c 1 1 1 1 1 1 1 1P . a

b c c a a b 3 b c c a a b 3 b c c a a b

3 3

2(a

b c

bb c) 2 (a c3 )

Li gii 2 : p dng BT Swcharz :

+ +

+ + + + += + +

+ + +

4 2 2 2 2

2 2 2 2 2 2 2 2 2

4 4a b c (a

P .a b c) b c a) c a b) b

b c )

( ( ( c ) a(b c ) c(a( ba )

Li c : + + + +

+ =

32 2 2 2 2 2 2

2 2 2a b c . b c 1 2aa(b2(b c )

c )32 2

42.( chn i tuyn QG d thi IMO 2005 ) . Cho a,b,c >0 . CMR : + + + + +3 3 3

3 3 3

a b c

(a b) (b c) (c a)

3

8

Li gii : = = ==b c a

x; y; z ; xyz 1a b c

. Bt ng thc cho tr thnh : + + + + +3 3 31 1 1 3

8(1 x) (1 y) (1 z)

p dng AM-GM ta c :( ) ( ) ( )

=+ + + +

+ + 363 3 2

11 1 13

81 x 1 x

3

8(1 x) 2 1 x

Ta cn CM bt ng thc : + + + + +2 2 21 1 1 3

4(1 x) (1 y ) (1 z)


12


13/46

B :( ) ( )

( )+ >++ +

2 2

1 1 1x,y 0

1 xy1 x 1 y

B ny c CM bng cch bin i tng ng a v BT hin nhin : + 2 2xy(x y) (1 xy) 0

Do :+ + + +

+ = + = =+ ++ + + + +

2

2 2 2 2

1 1 z 1 z(z 1) 1 z z 1

1 xy z 1(VT

1 z) (1 z) (1 z) z 2z 1

Gi s : = = 3z Max{x,y,z} 1 yz z zx 1 . Xt hm s :+ +

= + + +

=2 2

2 4

z z 1 z 1; f '(z) 0, z 1

z 2z 1 (z 1)f(z)

Suy ra : = 3f(f ) 1)(z4

.

43.( thi HSG Tnh H Tnh nm 2008 ) . Cho + + =0 :x yx y,z z, 1 . Tm gi tr nhnht ca :

+ ++ + +

=1 x 1 y 1 z

1 x 1P

y 1 z

Li gii 1 : ( ) +

+

22

2

1 x

1 x

x(1 x) 1 x 1 1 x 0 1 x 0

1 1 x( lun ng )

Thit lp cc BT tng t ta c : P 2

Ch : tm Max cn s dng BT ph :

+ + ++ + + +

1 x 1 y 1 x y1 , x y

1 x 1 y 1 x y

4

5v = +MaxP 1

2

3

44.( thi HSG lp 11 tnh H Tnh nm 2008 ) . Cho > + + =x,y,z 0 :x y z 1 . Chng minh bt ng thc : + + +

+ + + + + + +

1 x 1 y 1 z x y z2

y z z x x y y z x

Gii : BT

+ + + + + + + + + + + +

+

x y z x y z 3 xz xy yz2

y z z x x y y z x 2 y(y z) z(z x) x3

(x2

y)

Ta li c :( )+ +

= + + = + + + + + + + + + +

22 2 2 xz yz zxxz xy yz (xz) (xy) (yz)

VPy(y z) z(z x) x(x y ) xyz(y z) xyz(z x) xyz(x y ) 2xyz(x y z)

M :+ +

+ + = + + 2(xy yz zx)

xyz(x y z) (xy)(yz) (xz)(zy) (zx)(xy) VP3

3

2

45.( thi HSG Tnh Qung Bnh 2010 ) . Cho + + =0 :a ba b,c c, 3 . Chng minh rng : + + + + +3 3 31 1a c ab c 1b 5

46.Cho a,b,c l di 3 cnh tam gic ABC . Tm GTNN ca : = + ++ + + 2a 2b 2c

P2b 2c a 2a 2c b 2b 2a c

HD : = + + ++

2a 6a 6a

2b 2c a (a b c)(3a)(2b 2c a)

47.Cho + + =0: a ba b,c c, 1 . Tm GTLN, GTNN ca : + + += + + + ++2 2 2a 1 b 1P b c ca 1 HD .Tm GTNN : p dng BT Mincopxki ta c :

= + + = + +

+ + + + + + + + + +

2 22 2

2 2 2 1 3 3 3a 1 b 1 c 1 a a b c2 2 2

3P a b c

2

Tm GTLN :

B : CM bt ng thc : + + + + + + + + + +2 2 21 a a 1 b b 1 1 (a b) (a b)

Bnh phng 2 v ta c : + + + + + + + + + + + + + + 2 2 2 2(1 a a 1 a b (a b) 1 a b (a b))(1 (1 ab b ) b b)a 0

48.( thi chn HSG QG tnh Hi Dng nm 2008 ) . Cho > + + =a,b,c 0:a b c 3 . Tm gi tr nh nht ca biuthc : += +

+ + +

2 2 2

3 3 3

a b c

a 2b b 2c c 2aP

HD : AM-GM ngc du .

Ta c : = = + + = + +

2 3 33 2

3 3 3 6

a 2ab 2ab 2 2 2 4a a a b a a b(a a 1) a b ab

3 9 9 9a 2b a 2b 3 ab


13


14/46

Do :( )+ +

+ + + + = + + 2

a b c2 4 7 4(a b c) (ab bc ca) 1

9 9P (a b

9c

3)

3

49.( chn T trng chuyn Bn Tre ) . Cho x,y,z 0 . Tm GTLN ca : + + + + + +

=1 1

x y z 1 (1 x)(1 y)(1M

z)

Gii : t + + =x y z t 0 , ta c :+ + +

+ + +

3x y z 3

(1 x)(1 y)(1 z)3

. Lc : + + 31 2

M7

t 1 (t 3)

Xt hm s :

+ +

=3

1 27, t 0

t 1 (t 3

t

)

f( )

50.Cho >a,b,c 0 . Chng minh rng : + + + + ++ + + + +

4 4 4 2 2 23a 1 3b 1 3c 1 a b c

b c c a a b 2

HD : Ta c : + = + + + =44 4 4 4 1 31 a a a a3 4 2a 1 4a

Do : = + +

3 4

Svacxo

4a 4a...

b c ab acVT

51.Cho >a,b,c 0 . Chng minh rng : + + + + ++ + + + +

1 1 1 9 4 4 4

a b c a b c a b a c b c

HD :

52. Cho > + + =a,b,c 0 :a b c 1 . Chng minh rng :( ) ( ) ( )

+ +

+ + +

b c c a 3 3

4a 3c ab b 3a b c 3b ac

b

c

a

53.Cho >a,b,c 0 . CMR : + + + + + + + + + + 2 2 2 2 2 2 2 2 2a 1 1 1 1

6 a b c3a 2b c 3b 2c a 3c 2a b

b c

54.Cho > + + =a,b,c 0:ab bc ca 3 . CMR : + + + + +2 2 2a b c

abc2a bc 2b ca 2c ab

55. Cho >a,b,c 0 . CMR : + + ++ + + + +

3 3 3

2 2 2

1 a 1 b 1 c3

1 a c 1 c b 1 b a

56.Cho > =a,b,c 0:abc 27 . CMR : + + + + +

1 1 1 3

21 a 1 b 1 c

57.Cho >a,b,c 0 . CMR : + + + + + + + 2

1 1 1 27

b(a b) c(c b) a(a c) (a b c)

58.Cho >a,b,c 0 . CMR : + + ++ + + + +b c c a a b a b c 3a b c

59.Cho (a,b,c 1;2) . CMR : + +

c b

b c

b a a c1

4b c c a 4c ab a b4a

60.Cho > =a,b,c 0 : abc 1 .CMR : ++ + + +

3 6

a b c ab bc1

ca

61.Cho >x ,y ,z 0 . CMR : + + + + + + + 2 2 2

3 3 3

x z 1 x y z

2 y z xxyz y xyz z xyz x

y x z y

62.Cho + + => 1 1 1 1a

a,b,cb c

0: . CMR :+ +

+ + + + +

2 2 2a b c a b c

a bc b ac c ba 4

63.Cho >x ,y ,z 0 . Tm Min ca : = + + + + + + + +

3 3 3 3 3 33 3 3

2 2 2

x y zP 4(x y ) 4(y z ) 4(z x ) 2

y z x

64.Cho > + + =a,b,c 0: b ca 3 . CMR : + + + +a b c ab bc ca 65.Cho > =a,b,c 0:abc 1 . CMR: + +

+ + + + + +

1 1 11

a b 1 b c 1 c a 1

66.Cho >x ,y ,z 0 . CMR : + + + + + + + + + + +

x1

x (x y )(x z) y (x y )(y z) z (x z)(y z)

y z

67.( thi HSG Tnh Bnh Phc nm 2008 ). Cho >a,b,c 0 . CMR : + ++ + + + +

3 3 3

2 2 2 2 2 2

a b c a b c

2a b b c c a


14
http://d/HOC%20TAP/TAI%20LIEU%20BD%20HSG/CAC%20CHUYEN%20DE%20BDHSG/BDT_%20DAI%20SO/Bat%20Dang%20Thuc%20moi.pdfhttp://d/HOC%20TAP/TAI%20LIEU%20BD%20HSG/CAC%20CHUYEN%20DE%20BDHSG/BDT_%20DAI%20SO/Bat%20Dang%20Thuc%20moi.pdfhttp://d/HOC%20TAP/TAI%20LIEU%20BD%20HSG/CAC%20CHUYEN%20DE%20BDHSG/BDT_%20DAI%20SO/Bat%20Dang%20Thuc%20moi.pdfmailto:[email protected]


15/46

68.( thi HSG Tnh Thi Bnh nm 2009 ) .Cho cc s thcx , y , ztha mn + + =2 2 2x y z 3 . Tm gi tr ln nhtca biu thc: = + + + + +2 2F 3x 7y 5y 5z 7z 3x

69.( thi HSG TP H Ch Minh nm 2006 ) . Cho a,b,c l cc s thc khng m tha: + + =a b c 3 . Chng minh:+ +

+ + +

2 2 2

2 2 2

a b c 3

2b 1 c 1 a 1.

70. Cho a,b,c > 0 . Chng minh rng : + + + + +

2a 2b 2c3

a b b c c a

HD : t == = =a ;y ;z xb cxc a

yz 1b

. p dng B : ( )+ ++ +2 2

1 1 xy 11 xy1 x 1

2y

71.Chng minh c c Bt ng th c :a) ( )+ + ++ + >2 2 2b c c a a blog a log logb c 3 a,b,c 2 b) ( )+ + >

+ + + + +

b c alog c log a log 9

a ,b,c 1b c c a a b a b c

2

c)72. Cho + + =0:xyx,y yzz zx, 3 . Tm gi tr nh nht ca : += + + + + 22 3 2 3 23 22P x y (xy z z 1) (y 1) (z 1)x

Gii :73.


15


16/46

PHN IV : GII HNDY S

1. Cho dy s :( )

1

2

n 1 3 n

x 1

x 7 log x 11+

=

= +. Chng minh dy s c gii hn v tnh gii hn .

HD : Xt hm s : 23

f (x(x) 17 lo 1) 5g , x (0; )+ = , ta c :2

x (0;5)11)

2xf '(x) 0

n,

(x l 3

=

+<

Do : 0 f(5) f(x ) f(0) 5< < < < . Mn 1 n

f )x (x+ =

, do bng quy np ta CM c rng :n

, n0 x 5< <

Li xt hm s : 23(g( x 11) x , x (0;5)x) 7 log + = . Ta c :

2x (0;5)

11)l

2xg'(x) 1 0

(x 3,

n

= <

+

Suy ra phng trnh f(x)=x c nghim duy nht x = 4 .

Theo nh l Lagragen

(x 4)c ; sao cho : n n n1

f(x ) f(4) f '(c) x 4 x 411ln3

=

( V2 211)ln

2c 2c 1f'(c)

(c 11ln32 11c ln33= =

+). Do :

1

n 1 1

n

1x 4 x 0

11ln34

+

2. Cho phng trnh : 2n 1x x 1+ = + vi n nguyn dng . Chng minh phng trnh cho c duy nht mt nghimthc vi mi n nguyn dng cho trc. Gi nghim l

nx . Tm

nlimx

Gii : T phng trnh : 2n 2n2n 1 x 11) 1 1x x 1 x(x ) 0 x(x 1) 0x(x x 0+ > = > >


17/46

4. ( thi HSG Tnh Bnh nh nm 2010 ) . Cho dy s 1 2n n

n 1 n

1 2

x1

x

{x , n 1

2

x } :x +

= +

. Chng minh dy c gii hn v

tm gii hn . Li gii :

Bng quy np ta chng minh cn

0, nx 0> >

+) TH1 : Nu 0x 1= , quy np ta c n 1, nx 0= > . Hin nhin nlimx 1=

+) TH1 : Nu0

x 1> ,

Xt hm s :2

2

x(xf(x)

3)

13x

+

+= trn khong (1; )+ ta c :

2 2

2 2

xf '(x) 0

(x 1)x (1; ) f(x),

(3f( ) 1

x1

1)

+ >= > =

+

Do :2 1

) 1x f ,x .( .. .>= quy np ta c :n

x 1, n>

Li c : k k k kk k

k

2

2

k

2

k 1 2

(x 3) 1)x

1

x 2x (xx x 0

3x 3x 1+

+

<

++< > ng vi

kx 1>

T ta c :1 2 n n 1

x ....x x x 1+> > > >> . Dy s gim v b chn di nn tn ti giihn hu hn .

Gi s :( )2

n 2

a a 3

1

limx a 0 a a 1

3a

+=

+

> = =

+) TH3 : Nu0

0 1x< < , Xt hm s :2

2

x(xf(x)

3)

13x

+

+= trn khong (0;1) ta c :

2 2

2 2

(x 1)x (0;1) 0 f(0) f(

xf '(x) 0,

(x) f( ) 1

1)3x1

= < =

+

Do :2 1

f(x ) (0;1x ),...= quy np ta c :n

(0; nx 1),

ta c : k k k kk k

k

2

2

k

2

k 1 2

(x 3) 1)x

1

x 2x (xx x 0

3x 3x 1+

+

>

++> < ng vi

k0 1x< <

Do :1 2 n n 1

0 x x ... x x 1+< < < < < < . Dy s tng v b chn trn nn tn ti gii hn hu hn . Gi s :

( )2

n 2

a a 3

1limx a 0 a a 13a

+

= +> = =

Kt lun :n

limx 1=

10.( Bi ton tng t ) . Cho 0; a 0 > > l hai s ty . Dy 02n n nn 1 2

n

(u 3a)

a

u

{u } : uu ,n 0,1,...

3u+

=

+= =

+

. Chng minh dy

c gii hn v tm gii hn .

11.( Chn i tuyn H Vinh nm 2010 ) . Cho d y s0

2n n n

n 1

n

1

1 2(u 1

u

){u }: uu , n 0,1..

u 1.+

>

+ + +

= =

. Tmn

limu


18


19/46

12.( thi chn T HSG QG KonTum nm 2010 ) . Cho dy s thcn

{a } xc nh nh sau :1

n 1 n

n

1

1(na a )

a

1a

+

=

= +

Chng minh rng : nn

alim 2

n+=

13.( thi HSG Tnh Hi Dng nm 2006 ) . Cho dy s thc n1 n 1

2

n

x2006; x 3x

x 1+ +==

. Tm

nxlim x+

14.(thi HSG Tnh Ph Th nm 2008 ) . Cho dy s n{x } tha mn :1

n 1 n n n n

1

x (x 1)(x 2)(x 3

x

1 , 0x ) n+

=

+ +

+ + >=

. tn

i

n

i 1

1

xy

2= += . Tm nlimy .

HD : ( )2

2 2

n 1 n n n n n n n nn n n 1

1 1 1x (x 1)(x 2)(x 3) 1 x 3x 1 x 3x 1

x 2 x 1 1x

x+

+

+ + + + = + + = + = + +

= + +

Sau chng minh dy tng v khng b chn trn .

15.Cho dy 1n 2

n 1 n n

x a 1):

2010x(

0 9xx

2 0 x+

= >

= +. Tm : 1 2 n

n2 3 n 11 1 1

x x xlim ...

x x x ++

+ + +

HD : Xt hm s :2x 2009x

f(x) , x 12010 2010

= + > . Ta c : f(x) > 0 , x 1 > f (x) f (1) 1 > = . Bng quy np chng minh

c rng :n

x 1, n> . Xt hiu :2n n n n

n 1 n n n 1 n

x x x (xx 0,

2010 2010 201

1)x

0x 1 x x+ +

> = = > >

Gi s ( ) 2nlimx a a 1 201 2009a a 0;a 10a a = > = + = = ( Khng tha mn ). Vy nlimx = + Li c :

2 n n 1 n

n 1 n n n 1 n n n

n 1 n n 1 n n 1

x 1 12010x x ) x 1) 2010 2010

x 1

x x2009x 2

(x 1)(x 1) x 1 x 1010(x x (x ++ +

+ + +

= = = =

+

16.( Bi tng t ) . Cho dy s : 124n n

n 1 n

x 1

): xx x N *

2

(x, n

4+

=

= +

. Tm gii hn23 23 23

1 2 n

2 3 n 1

x x xlim ...

x x x +

+ + +

17.( thi HSG Tnh Bnh Phc nm 2008 ) . t 2 2f(n) (n n 1) 1+ + += vi n l s nguyn d ng .Xt dy sn n

f(1).f(3).f(5)...f(2n 1)(x

f(2).f(4).f(6)...f nx

)):

(2

= . Tnh gii hn ca dy s : 2n n

u n .x=

HD : Ch :2

2

f(k 1) (k 1)

f(k) (k 1

1

1)

+

+

+=

18. Chody sn

(a ) xc nh bi : n

i 1

1

2

i n

2a

a n

008

,n 1a=

= >

=

. Tnh 2n

nim al n+

HD : Ta c ( ) ( )22 2

1 2 n n n 1 n n n 1

n 1a n n 1 a n a aa ... a a

11 a

n

= = =

++ + + (1)

Trong (1) cho n=1,2,3.v nhn n l i tm : an

19.Cho dy s (

nx ) tha : 1 n 1n

2006x 1,x 1 (n 1)1 x

+= = + + .Chng minh dy s ( nx ) c gii hn v tm gii hn y

20.( thi HSG QG nm 2009 ) . Cho dy s 1n 2

n 1 n 1 n 1

n

1x

2):

x 4x xx

2

(

, n 2

x

=

+ + =

. Chng minh rng dyn

(y ) vi

n 2

n

i 1 i

1y

x== c gii hn hu hn khi n v tm gii hn .

Gii :


19
http://d/HOC%20TAP/TAI%20LIEU%20BD%20HSG/DE%20THI%20HSG/DE%20THI%20HSG%20LOP%2012/DT&DA%20HSG%20PHUTHO_08-09.pdfhttp://d/HOC%20TAP/TAI%20LIEU%20BD%20HSG/DE%20THI%20HSG/DE%20THI%20HSG%20LOP%2012/DT&DA%20HSG%20PHUTHO_08-09.pdfhttp://d/HOC%20TAP/TAI%20LIEU%20BD%20HSG/DE%20THI%20HSG/DE%20THI%20HSG%20LOP%2012/DT&DA%20HSG%20PHUTHO_08-09.pdfhttp://d/HOC%20TAP/TAI%20LIEU%20BD%20HSG/DE%20THI%20HSG/DE%20THI%20HSG%20LOP%2012/DT&DA_HSG%20BINHPHUOC%20V1_2_08-09.pdfhttp://d/HOC%20TAP/TAI%20LIEU%20BD%20HSG/DE%20THI%20HSG/DE%20THI%20HSG%20LOP%2012/DT&DA_HSG%20BINHPHUOC%20V1_2_08-09.pdfhttp://d/HOC%20TAP/TAI%20LIEU%20BD%20HSG/DE%20THI%20HSG/DE%20THI%20HSG%20LOP%2012/DT&DA_HSG%20BINHPHUOC%20V1_2_08-09.pdfhttp://d/HOC%20TAP/TAI%20LIEU%20BD%20HSG/DE%20THI%20HSG/DE%20THI%20HSG%20LOP%2012/DT&DA%20HSG%20PHUTHO_08-09.pdfmailto:[email protected]


20/46

Xt hm s :2

x xf(x)

2

4x+ += , ta c :

2

2x 4 1f '(x) 0,

24x

4xx0 >

+

+= + >

Li c :2 1 1

f(x ) 0,(do x 0)....x = > > bng quy np ta chng minh cn

x 0, n> .

Xt hiu :2 2

n 1 n 1 n 1 n 1 n 1 n 1 n 1

n n 1 n 1 n2

n 1 n 1 n 1

x 4x x x 4x x 4xx x 0,(do x

2x 0,

xn

x 4)

2 x

+ + + = = = >>

+

+

Suy ra dyn

{x } tng vn

x 0, n> . Gi s tn ti gii hn hu hnn

n0im ( )a l x a

+

= > . Suy ra :

22a aa a a a 0

24a 4a+= += + = (V l ) .

Vy dyn

{x } tng v khng b chn trn nn :n

nlimx

+

= +

Li c :

( )2

2n 1 n 1 n 1 2 n n n 1 n 1

n n n 1 n 1 n 1 n n n 1 n 1 2 2 2

n 1 nn n 1 n n 1 n

x 4x x x (x x ) xx 2x 4x x (x x )

1 1 1x x

x xx .x xx

2 .x x

+ + = = + = = =

Do : 1n n2 2 2

1 2 n 1 n ni 1

n

ni 1 1

1 x1 1 1 1 1 1 1y ... lim y 6

x x x x xx x x +

=

+= = + + + = =

.

21.Xt dy s thc n(x ), n N xc nh bi : 03n n 1 n 1

2009

6x 6sin(x

x

), n 1x

=

=

. Chng minh dy c gii hn hu hn

v tm gii hn .

HD : S dng bt ng thc :3

xx ins x,x

6x 0

Xt hm s : 3f(x) 6x 6 sin x , x 0= > . Ta c :3 2

1 6(1 cosx)f '(x) 0, x>0

3 (6x 6sin x)

= >

Do : f(x) 0 x, 0> > . M2 1 1 n n 1

f(x ) 0(do x 0) ...x f(x ) 0, nx = > > = >

Xt hiu : 33

3

n 1 n 1 n 1

n n 1 n 1 n 1 n 12 2

n 1 n 1 n 1 n 1 n 13

n 1

)x 6sin(x

6x x 6sin(xx 6x x 0)

6sin(x ) 6sin6x x 6x x(x )

= = , xt : 2n n

2

n

2

n n

n 1 n

n

n n

n

8x 11x 3x x 9x +11x + 3 x 0,

9x +11x + 3x 0

x+

+ + = > =

+>

Gi s ( )n

2

n

a 1lim x a a 0 a 9a 1a 3 3

a8

1+

+=

= > = =

+

( Khng tha mn ) n

nlim x+

= +

Do :n n

n 1

2

n n n

x 11 3lim lim 9 3

x x x+ +

+= + + =


20


21/46

24.Cho dy sn

(u ) xc nh bi cng thc

1

2 2

n+1 n n

u = 2008

u = u - 4013u + 2007 ; n 1, n N.

a)Chng minh:n

u n + 2007; n 1, n N .

b)Dy s (x n

n

1 2 n

1 1 1x = + + ... + ; n 1, n N.

u - 2006 u - 2006 u - 2006

) c xc nh nh sau:

Tm nlimx ?

25.( thi HSG Tnh Tr Vinh-2009)Cho dy s (n

U ) xc nh bi:1

33n 1 3 n

U 1

4U log U 1 , n 1

3+

=

= + +

Tmn

n

limU+

26. Cho dy s ( )nn

0

xn n

n 1 x

x 1

): 2 l(x x 1

ln 2 1

n2 1x

2+

=

+

=

. Chng minh dy (xn

HD : Chng minh dy gim v b chn di .

) c gii hn v tm gii hn .

27. Cho phng trnh : n n 1x x .... x 1 0+ + + = . Chng t rng vi n nguyn dng th phng trnh c nghim duynht dng

nx v tm

nxlim x+

.

28.Cho dy sn

{u } xc nh bi1

n n

n 2n

1

C n

u

u .4 =

=

. . Tmn

limu

29.( thi HSG Tnh Ngh An nm 2008 ) . Cho ph ng trnh:x

1x n 0

2008 + = (1). Chng minh rng: vi mi n

N*phng trnh (1) c nghim duy nht, gi nghim l x n . Xt dy (xn), tm lim (xn + 1 - xnp n :

).

Vi n N*x

1x n

2008 +, xt f (x) = ; x R.

f/x

ln2008

2008(x) = - - 1 < 0 x R.

=> f(x) nghch bin trn R (1).

Ta c:n

n 1

1f(n) 0

2008

1f(n 1) 1 0

2008 +

= >

+ = f(x) =0 c nghim x n (n; n + 1) (2). T (1) v (2) => pcm.

Ta c: xnnx

1

2008- n = > 0 => xn > n.

=> 0 < xnn

1

2008- n < .

Mt khc: lim n1 02008 = => lim(xn - n) = 0.

Khi lim (xn - 1 - xn) = lim{[xn + 1- (n + 1)] - (xn - n) + 1} = 1

MT S BI TON TM GII HN KHI BI T CNG THC T NG QUT C A DY S .

30.Chody s ( ) 1n n 1n

n 1

2

24

u

u : 9uu , n 2

5u 13

=

+

=

. Tmn

limu ?=

Gii :


21


22/46

31.Cho dy s ( ) 1n2

n n 1

1

2

2

uu :

u u 1 , n 2

=

=

. Tm nn

ulim

n+

HD: Tm c :n 1

n

2u cos

3

=

v ch :x

n nu u1

0 lim 0n n n+

=

32.Cho dy s ( )1

n 2

n 1

n

u

u :

2 2 1 uu2

1

2

, n 2

=

=

. Tm nn

nlim 2 .u+

HD : Tm cn n 1

u sin2 .6

= suy ra : n

n

n n

n

n

sin3.2

lim 2 .u lim3 3

3.2

+ +

= =

33.Cho dy s ( )1

n 1nn

2

n 1

u

u :u

1 1

3

u, n

u2

+ +

=

=. Tm n

nnlim 2 .u+

HD : Tm cn n 1

u tan

3.2

=

34.Cho dy s ( ) 1nn 1

n

n 1

2

3

u, n 2

2(2n 1

u :

u)u 1

u

=

=

+

. Tmi

n

ni 1

lim u+

=

35.Cho dy s : 12

n 2 n n 1

1

2

u 2u N *

u

u

u , n+ +

=

=

= +

. Tm n 1

nn

ulim

u++

HD : Tm c ( ) ( )n

n

n2u 1 1

42 2= +

. Suy ra :

( ) ( )

( ) ( )( )

n 1

n 1

n 1

n 1

x nn nn

1

2

2 2 2

12 1

1 1 1u 4lim

u 21 1 11 1

41 11

2 1

22 2

2 22

+

+

+ +

+

+

+ + = = = + + +

+

36.Cho dy s ( )1

n n 1

n

n 1

u

u :u

1 3

3

3 u, n 2

u

+=

=

. Tnh nn

ulim

n+

HD :n

nu tan

3

=

37.Cho dy sn

(u ) xc nh nh sau : nu 2 2 2 .... 2= + + + + ( n du cn ) . Tnh1 n

nn

2u .u ...u

lim2+

HD : t :n

n

n n 1

ux x cos

2 2 +

= = v ch : 1 2 n 1 2 nn nn 1

sinu .u ...u 1 2x ...x

2 2si

.

2

x

n+

= =


22


23/46

38.Cho dy s 1n

2

n 1 n n n

1b

2):

1 1b b

(b

b (n 1)2 4

+

=

+

+

=

. Chng minh dy hi t v tmn

nlim b+

HD : Chng minh :n n n 1

1b .cot

2 2 +

=


23


24/46

PHN V : HNH HC KHNG GIAN

1. Cho hnh chp tam gicu c th tch l 1. Tm gi tr ln nht ca bn knh mt cu ni tip hnh chp. 2. Cho t din ABCD c : AB=a; CD=b ; gc gia AB v CD bng .Khong cch gia AB v CD bng d. Tnh th tch

khi t din ABCD theo a,b,d v .3. Trong cc t din OABC c OA, OB, OC i mt vung gc vi nhau v th tch bng 36. Hy xc nh t din sao

cho din tch tam gic ABC nh nht.

4. Cho hnh hp ABCD.A 1B1C1D 1 . Cc im M, N di ng trn cc cnh AD v BB 11

MA NB

MD NB=sao cho . Gi I, J ln

lt l trung im cc cnh AB, C1D 15. Gi O l tm ca mt hnh t din u . T mt im M bt k trn mt mt ca t din , ta h cc ng vung gc

ti ba mt cn li. Gi s K, L v N l chn cc ng vung gc ni trn. Chng minh rng ng thng OM i quatrng tm tam gic KLN.

.Chng minh rng ng thng MN lun ct ng thng IJ.

6. Cho hnh chp S.ABC . T im O nm trong tam gic ABC ta v cc ng thng ln l t song song vi cc cnhSA, SB, SC tngng ct cc mt (SBC), (SCA), (SAB) ti cc im D,E,F .

a) Chng minh rng : OD DE DF 1SA SB SC

+ + =

b) Tm v tr ca im O trong tam gic ABC th tch ca hnh chp ODEF t gi tr ln nht. 7. Cho hnh hp ABCD.A1B1C1D 1 . Hy xc nh M thuc ng cho AC1 v im N thuc ng cho B1D 1 ca mt

phng A1B1C1D 1 sao cho MN song song vi A 1

8.

Cc im M, N ln lt l trung im ca cc cnh AC, SB ca t din u S.ABC . Trn cc AS v CN ta chn ccim P, Q sao cho PQ // BM . Tnh di PQ bit rng cnh ca t din bng 1.

D.

9. Gi O l tm mt cu ni tip t din ABCD. Chng minh rng nu 0ODC 90= th cc mt phng (OBD) v (OAD)vung gc vi nhau .

10. Trong hnh chp tam gicu S.ABC (nh S ) di cc cnh y bng 6 . di ng cao SH = 15 . Qua B vmt phng vung gc vi AS, mt phng ny ct SH ti O . Cc im P, Q tng ng thuc cc cnh AS v BC sao

cho PQ tip xc vi mt cu tm O bn knh bng2

5. Hy tnh di b nht ca on PQ.

11. Cho hnh lp phng ABCD.A 1B1C1D 1 cnh bng a . ng thng (d) i qua D 1 v tm O ca mt phng BCC 1B1 .on thng MN c trung im K thuc ng thng (d) ; M thuc mt phng (BCC1B1

12. Cho t din ABPM tho mn cc iu kin :) ; N thuc mt y (ABCD) .

Tnh gi tr b nht ca di on thng MN . 0 2AM BP; MAB ABP 90 ; 2AM.BP AB = = = . Chng minh rng mt

cu ng knh AB tip xc vi PM. 13. ( thi HSG Tnh Qung Ninh nm 2010 ) Cho im O c nh v mt s thc a khng i . Mt hnh chp

S.ABC thay i tha mn : OA OB OC a; SA OA;SB OB;SC OC= = = ; 0 0 0ASB 90 BSC 60 CSA; ; 120= = = . Chngminh rng :

a. ABC vung . b. Khong cch SO khng thay i .

Gii :a) t : SO = x .

Ta c : Cc tam gic OAS, OBS, OCS vung nn : 2 2SA SB SC ax= = = .Do : 2 2 2 2 2AB S SB a )A 2(x= =+ ; 2 2 2 0 2 2SC 2SA.SCAC SA os120 3(x. a )c= =+ ;

2 2 2 0 2 2SB SC 2SB.SBC os6C.c a )0 (x= + = 2 2 2AB BCAC = + hay tam gic ABC vung ti B.

b) Gi M l trung im AC , do cc tam gic SAC, OAC l cc tam gic cn nn :SM AC

AC (SOM) AC OSOM AC

Tng t, gi N l trung im AB, ta CM c : AB SO Suy ra : SO (ABC) .Do mi im nm trn ng thng SO ucch u A, B, C . Suy ra SO i qua tm ngtrn ngoi tip M ca tam gic A BC . Trong cc tam gi c vun g ABC v SBO ta c h


24


25/46

thc :2 2 2

2 2 2

1 1 1

BM AB BC

1

BM

1 1

OB BS

= +

=

+

2 2 2 2

1 1

OB BS

1 1

AB BC + = + 2 2

2 2 2 2 2 2 2

1 1 1 1 33a x a

22(x x a2x

) axa a + = + = =

14.( thi HSG Tnh Vnh Phc nm 2010 ) . Cho hnh chp S.ABCD cy ABCD l hnh ch nht , AB = a ;BC 2a= . Cnh bn SA vung gc vi y v SA=b . Gi M l trung im SD, N l trung im AD .

a) Chng minh AC vung gc vi mt phng (BMN) b) Gi (P) l mt phng i qua B, M v ct mt phng (SAC) theo mt ng thng vung gc vi BM .

Tnh theo a, b khong cch t S n mt phng (P) . Li gii :

t AS x;AB y;AD z x.y y.z z.x 0;| x| b;| y | a;| z | a 2= = = = = = = = = Ta c : AC AD AB y z= + = +

v

1BN AN AB z y

2= =

Do :2

2 2 2(ay a 0 A1 2)

AC. C BNN z2 2

B = = =

Li do : 1MN SA MN AC2

=

Hay : AC (BMN) AC BM Gi s (P) ct (SAC) theo giao tuyn (d) BM M do (d) v AC ng phng (d)//(AC)

Gi O (AC) (BD)=

Trong mt phng (SDB) : SO ct BM ti I. Qua I k ng thng (d) // (AC) ct SA, SC ln ltti H, K . Mt phng (MHBK) l mt phng (P) cn dng . Li v : I l trng tm tam gic SDC v HK//AC nn :

SH SK SI 2

SC SA SO 3= = = (1)

Theo cng thc tnh t s th tch ta c :

SMBK SMHB

SDBA SDCB

V VSM SB SK 1 SM SH SB 1. . ; . .

V SD SB SA 3 V SD SC SB 3= = = =

2

SABCD

SKMHB SKMB SMHB SDBA

V2 b 2V

aV V V

3 3 9 = =+ = = (2)

Ta li c :KMHB MKH BKH

1 1 1S MI.HK BI.HK BM.HKS

2S

2 2= + == + (3)

M : 2 22 2 2

HK AC a3.a

(a 2)3 3 3

+= = = ; ( ) ( )1 1

BM AM AB AS AD AB x z y2 2

= = + = +

( )2 2

2 2 2 2 2 2 6aBM) z1 1 3 b

( x y b a BM4 4 2 2

= + =+

+ =+

(4)

T (3), (4) suy ra :2 22 2

KMHB

a 3(b1 b 2 6a )6S .

2

a 3a

2 3 6

++= = (5)

T (2), (5) suy ra :2

SKMHB

2 2 2 2KMHB

b 23V 18a 2d(S,(P))

S 9a. 3(

2ab

6a 6ab ) 3(b )+ += = =

15.( thi HSG Tnh Bnh Phc nm 2010 ) . Cho hnh lp phng ABCD.ABCD c cnh bng a . Trn AB lyim M, trn CC ly im N , trn DA ly im P sao cho : AM CN D'P x x a)(0= = = .

a) CMR tam gic MNP l tam gic u, tm x din tch tam gic ny nh nht . b) Khi ax

2= hy tnh th tch khi t din BMNP v bn knh mt cu ngoi tip t din .


25


26/46

16.( thi HSG Tnh B Ra Vng Tu nm 2008 ) . Cho t din ABCD c cc cnh AB=BC=CD=DA=a ,AC x; BD y= = . Gi s a khng i, xc nh t din c th tch ln nht.

17.( thi HS G Tnh B Ra Vng Tu nm 2009 ) Cho khi t din ABCD c th tch V . im M thuc min trongtam gic ABC . Cc ng thng qua M son g song vi DA, DB, DC theo th t ct cc mt phng (DBC), (DCA),(DAB) tng ng ti A 1 ; B1 ; C 1

a) Chng minh rng :.

1 1 1MA MB MC

1DA DB DC

+ + =

b) Tnh gi tr ln nht ca khi t din1 1 1

MA B C khi M thay i .

18.( thi HSG Tnh Hi Phng nm 2010 ) . Cho t din OABC c OA, OB, OC i mt vung gc . Gi ; ; lnlt l gc to bi cc mt phng OBC, OAC, OAB vi mt phng (ABC ).

a) Chng minh rng : 2 2 2 2 2 2tan tan tan 2 tan .tan .tan + + + = b) Gi s OC=OA+OB . Chng minh rng : 0OCA OCB ACB 90+ + =

19.( thi HSG Tnh Ngh An nm 2008 ) .Cho t din ABCD c AB = CD, AC = BD, AD = BC v mt phng (CAB)vung gc vi mt phng (DAB). Chng minh rng:

1CotBCD.CotBDC = .

2

Li gii : t : BCD ; BDC= = Ta c :

BAC BDCABC DCB

ABC BCD

= = =

= =

BAD BCDCBD ADB

ABD CDB

= = =

= =

Gi H l hnh chiu ca C ln AB . t HC x= .

DoCBA DAB

CH DH(CBA) (BDA)

=

Trong tam gic vung BHC :HC HC x

sin BC ADBC sin sin

= = = =

HC x xtan BH

BH BH tan = = =

.

Trong tam gic vung AHC :

HC HC x

sin AC BDAC sin sin = = = = .

HC x xtan AH

AH AH tan = = =

Trong tam gic BCD : ( ) ( )2 2

2 2 2

2 2

x x x xCD BC os 2 . cos

sin sinsiBD 2BC.BD.c

n sin= = ++ + +

(1)

Li c :2 2 2H AD 2AHD .AH oc sAD.+ =

2 22

2 2

x x x xHD 2 . .cos

tan sintan sin +

= (2)

M tam gic CHD vung nn:

2 2(1) (

22)

CH HDCD+

= + ( )2 2 2 2

2

2 2 2 2

x x x x x x x x2 . cos x 2 . .cos

sin sin tan sinsin sin tan sin+ + + = +

+

2 22 2 1(1 cot ) (1 cot ) 2(cot .cot 1) 1 cot (1 cot ) 2cot .cot cot .cot 2

+ + + + = + + + =

P N CA S GD&T :


26


27/46

t AD BC a ,AC BD b, AB CD c ,BAC A ,ABC B, ACB C.= = = = = = = = =

Ta c ABC nhn v ABC = DCB = CDA = BAD.

Suy ra ( )BCD ABC B; ABD BDC CAB A, 1= = = = =

H CM AB ,v ( ) ( )CAB DAB nn ( ) ( )2 2 2CM DAB CM MD CM DM CD , 2 . + =

p dng nh l cosin cho tam gic BMD ta c ( )2 2 2MD BM BD 2BM.BD.cosMBD, 3= +

T (1), (2), (3) ta c 2 2 2 2CM BM BD 2BM.BD.cosA CD+ + = 2 2 2 2 2 2BC BD 2BM.BD.cosA CD a b 2abcosA.cosB c+ = + =

1cosC cosA.cosB sinA.sinB 2cosA.cosB cotA.cotB .

2 = = =

20.( thi HSG Tnh Ngh An nm 2008 ) .Cho khi ch p S. ABCD c y ABCD l h nh bnh hnh. Gi M, N, P lnlt l trung im ca cc cnh AB, AD, SC. Chng minh rng mt phng (MNP) chia khi chp S.ABCD thnh haiphn c th tch bng nhau.

21.( thi HSG Tnh Ngh An nm 2009 ) . Cho tam gic ABC , M l mt im trong tam gic ABC. Cc ngthng qua M song song vi AD, BD, CD tng ng ct cc mt phng (BCD), (ACD) , (ABD) ln lt ti A, B, C .Tm M sao cho MA'.MB'.MC' t gi tr ln nht.

Li gii 1 : tDABC MABD MAC BDC MBC AA B C

V ; ;V; V V V V V V V VV V= = = + += = v :

DA a; BD b; DC c; M A' x;MB' y ;MC ' z= = = = = =

Ta c : CV d(C,(ADB)) MC' z

V d(M,( ADB)) CD c= = = ; tng t : A B

V Vx y x y z; 1

V a V b a b c= = + + =

p dng bt ng thc AM-GM : 3x y z xyz abc

1 3 xyza b c abc 27

= + + . Du = xy ra

x y z 1

a b c 3 = = =

Do : MA'.MB'.MC' t gi tr ln nht khi v ch khi M l trng

Tm tam gic ABC .

Li gii 2 : t : DA a; BD b; DC c; M A' x ;MB' y ;MC' z= = = = = =

Ta c :A'M x

A M .DA DADA a

= =

;B M y

B M .DB .DBDB b

= =

;

P N S GD&T :


27


28/46

Trong mt phng (ABC) :AM BC = {A1}; BM AC = {B1}, CM AB = {C 1}

Trong (DAA1K ng thng qua M song song vi AD ct DA

) :1 ti A

Xt tam gic DAA1 MBC1

1 ABC

SMAMA'

DA AA S

= =c MA // AD nn

Tng t ta c MAC11 ABC

SMBMB'DB BB S

= = , 1 MAB

1 ABC

MC SMC'DC CC S

= =

Suy ra ( )MBC MAC MAB ABCMA' MB' MC'

1 doS S S SDA DB DC

+ + = + + =

Ta c 3MA ' MB' MC' MA ' MB' MC'

3 . .DA DB DC DA DB DC

+ +

Suy ra MA.MB.MC 1

27DA.DB.DC (khng i)

Vy gi tr ln nht MA.MB.MC l1

27DA.DB.DC, t c khi

1 1 1

1 1 1

MA MB MCMA' MB' MC' 1 1

DA DB DC 3 AA BB CC 3= = = = = =

Hay M l trng tm tam gic ABC

22.( Tp ch THTT : T10/278 ; T10/288 ) . Cho t din S.ABC vi SA=a; SB =b ; SC = c . Mt mt phng ( ) thay ii qua trng tm ca t din ct cc cnh SA, SB, SC ti cc im SA, SB, SC ti cc im D, E, F tng ng .

a) Tm gi tr nh nht ca cc biu thc :2 2 2

1 1 1

SD SE SF+ +

b) Vi k : a=b=c=1, tm gi tr ln nht ca : 1 1 1SD.SE SE.SF SF.SD

+ +

Li gii : t : SD x; SE y ; SF z= = =

G l trng tm t din nn : ( )1 1 SA 1 a

SG SA SB SC .SD .SD4 4 SD 4 x

= + + = =

Do D,E,F, G ng phng nn :a b c

4x y z

+ + = . T ta c :

( )2

2 2 2

2 2 2 2 2 2 2 2 2

1 1 1 a b c 1 1 1 16a b c (1)

x y zx y z x y z a b c16

+ + + + + + + + + +

=

Du bng xy ra

2 2 2

2 2 2

2 2 2

a b c

4a

a b c

4

x

y

z

b

a b c4c

+ +

+ +

+ +

=

=

=

23.( thi HSG Tnh Ngh An nm 2009 ) . Cho t din ABCD c di cc cnh bng 1 . Gi M, N ln lt l trungim ca BD, AC . Trn ng thng AB ly im P , trn DN ly im Q sao cho PQ song song vi CM . Tnh diPQ v th tch khi AMNP . Li gii :

Gi s : AB x; AC y; AD z= = =

v :AP

m;AQ n.AC (1 n)ADAB

= = +

Ta c :1

x.y y .z z.x2

= = =

Lc :


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( )1 n

AC y; AM x z ; AP m.x; AQ n.AN (1 n )zAD .y (1 n )z2 2

= = + = = + = +

Suy ra : ( )1

CM AM AC x 2y z2

= = +

nPQ AQ AP mx y (1 n)z

2= = + +

Do CM // PQ nn :

km

2

n 2PQ kCM k k 2 3

k1 n

2

=

= = =

=

Vy : ( ) ( )221

PQ 2y x z |PQ1 1

| 2y x z3

3PQ

9 3 3= = = =

P N CA S GD&T :

Trong mt phng (ACM) k NI // CM (I AM)Trong mt phng (BCD) k BK // CM (K CD)Trong (ABD) DI ct AB ti P Trong (AKD) DN ct AK ti Q

PQ l giao tuyn ca (DNI) v (ABK) ,do NI // CM, BK // CM nn PQ // CM

Gi E l trung im PB, ME l ng trung bnh tam gic BPD nn ME // PD hay ME // PI Mt khc t cch dng ta c I l trung im AM nn P l trung im AE.Vy AP = PE = EB

Suy ra AP 1AB 3

=

MC l ng trung bnh tam gic DBK nn BK = 2CM = 3

Suy raPQ AP 1

BK AB 3= = PQ =

1

3BK =

3

3

AMNP

AMCB

V AM AN AP 1 1 1. . .

V AM AC AB 2 3 6= = =

VAMCB1

2 = VABCD (Do M l trung im BD)

ABCD l t din u c di cnh bng 1 nn V ABCD2

12

= (vtt)

Suy ra VAMCB1 2 2

.2 12 24

== . Vy VAMNP1

6= V AMCB

2

144= (vtt)

24.( d b khi D 2008 ) . Cho t din ABCD v cc im M, N, P ln lt thuc cc cnh BC, BD, AC sao choBC 4BM; AC 3AP; BD 2BN= = = . Mt phng (MNP) ct AD ti Q . Tnh t s

AQ

ADv t s th tch hai phn ca khi

t din ABCD c phn chia bi (MN P).Li gii :

t : AB b;AC c; AD d= = =

Ta c :


29


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( )1

AN b d (2)2

= +

1AC 3AP AP c (3)

3= =

Do C,D,I v M, N, I thng hng nn :

AI mAC (1 m)AD 3 1 1 1c (1 m)d n b c (1 n) b d

4 4 2 2AI n AM (1

m

n)AN

= + + = + + +

=

+

n

ID 14 2AI 3AD AC2DI CD1 n IC 3

AI 2AM 3AN2 IN 2NI 2M

m

n 2

1 m 1m

2c d

N1 3

3n 1 n AI IM 30 2 2

4 2

= = = = +

=

=

= =

= =

= + + =

Gi s : AQ kAD=

. Do P, Q , I thng hng nn :

3p 1 pp

p 1 3 53 2AQ pAP (1 p)AI kd c (1 p) c d 5AQ 3AP 2AI 3PQ 2QI3 2 2 33(1 p)

kk 52

== = + = + + = + =

==

Suy ra :QI 3

PI 5=

Ta li c : IQND QPMCDN

IPMC IPMC

V VIQ IN ID 3 2 1 2 13. . . .

V IP IM IC 5 3 3 15 V 15= = = = (4)

M :( )( )

BCDABCD

PMCI MIC

d A,(BCD) .SV AC CB.CD.sinC 3 4 2 4. . .

V PC MC.CI.sinC 2 3 3 3d P,(MIC) .S= = = = (5)

T (4) v (5) suy ra : PQDNMC PQDNMC

ABCD ABMPQN

V V13 3 13 13

V 15 4 20 V 7= = =

25.( thi HSG Tnh H Tnh nm 2008) . Cho hnh chp t gic u S.ABCD c gc gia mt bn v y l . Vng cao SH ca hnh chp, gi E l im thuc SH v c khong cch ti hai mt phng (ABCD)v (SCD) bngnhau . Mt phng (P) i qua E, C, D ct SA, SB ti M, N .

a) Thit din l hnh g ?b) Gi th tch cc khi t din S.NMCD v ABCDNM ln lt l V 1 , V 2 . Tm 3V2 =5V1

26.( thichn T HSG QG tnh Qung Bnh nm 2010 ) . Cho t din ABCD . Gi trung im ca AB, CD ln ltl K , L . Chng minh rng bt k mt phng no i qua KL u chia khi t din ny thnh 2 phn c th tch bngnhau.

.

27.( thi HS G Thnh Ph Cn Th nm 2008 ) . Trong khng gian cho hnh chp S.ABC , trng tm ABC l G .Trung im ca SG l I . Mt phng ( ) i qua I ct cc tia SA, SB, SC ln lt ti M, N, P ( Khng trng vi S ) . Xc

nh v tr ca mt phng ( ) th tch khi chp S .PMN l nh nht .

28.( thi HSG Tnh Hi Dng nm 2008 ) .Cho hnh lp phng1 1 1 1BAB .A CD DC cnh bng 1 . Ly cc im M,

N, P, Q, R , S ln lt thuc cc cnh AD, AB, BB1 , B 1C1 , C1D 1, DD 1

29.Cho hnh chp t gic S.ABCD ,c y ABCD l mt hnh bnh hnh. Gi G l trng tm ca tam gic SAC. M l mtim thay i trong min hnh bnh hnh ABCD .Tia MG ct mt bn ca hnh chp S.ABCD ti im N .

. Tm gi tr nh nht ca di ng gpkhc khp kn MNPQRSM .

t: Q =MG NG

NG MG+

a) Tm tt c cc v tr ca im M sao cho Q t gi tr nh nht .b) Tm gi tr ln nht ca Q.

30.Trong mt phng (P) cho tam gic ABC . Ly im S khng thuc (P) . Ni SA, SB, SC . I l mt im bt k trongtam gic , gi AI ct BC ti A 1 , CI ct AB ti C1 , BI ct AC ti B 1 . K IA2//SA, IB2//SB, IC2

( )2 2 2(SBC);B (SAC);A C (SAB)

//SC

. CMR : 2 2 2

1 2 1 2 1 2

SA SB SC6

A A B B C C+ +

( ) ( )3 1

BC 4BM AC AB 4 AM AB AM b c (1)4 4

= = = +


30


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32/46

50.Cho t din ABCD c AB vung gc vi AC v chn ng vung gc h t A n mt phng (BCD) l trc tm tamgic BCD . Chng minh rng : ( ) ( )

2 2 2 26 AB AD AB CD DB CC + ++ +

51.( thi HSG TP H Ni nm 2004 ) . Cho t din ABCD DA=a, DB=b, DC=c i mt vung gc vi nhau.Mt imM tu thuc khi t din.

a) .Gi cc gc to bi tia DM vi DA, DB, DC l , ., .CMR : 2 2 2sin sin sin 2 + + = b) .Gi

A B C DS ,S ,S ,S ln lt l din tch cc mt i din vi nh A, B, C, D ca khi t din. Tm gi tr

nh nht ca biu thc:A B C D

Q MA.S MB.S MC.S MD.S= + + +

52.( thi HSG TP H Ni nm 2005) .Hnh chp S.ABC c cc cnh bn i mt vung gc v SA =a, SB=b, SC=c.Gi A, B, C l cc im di ng ln lt thuc cc cnh SA, SB, SC nhng lun tha mn SA.SA =SB.SB=SC.SC. GiH l trc tm ca tam gic ABC v I l giao im ca SH vi mt phng (ABC). a) Chng minh mt phng (ABC) song song vi mt mt phng c nh v H thuc mt ng thng c nh. b) Tnh IA2+IB2+IC2

53.( thi HSG TP H Ni nm 2006) .Cho t din u ABCD c cnh bng 1. Cc in M, N ln lt chuyn ngtrn cc on AB, AC sao cho mt phng (DMN) lu n vung gc vi m t phng (ABC). t AM=x, AN=y.

theo a, b, c.

a) . Cmr: mt phng (DMN) lun cha mt ng phng c nh v : x + y = 3xy. b) . Xc nh v tr ca M, N din tch ton phn t din ADMN t gi tr nh nht v ln nht.Tnh cc gi tr .

54.( thi HSG TP H Ni nm 2008 ) . Cho hnh chp S.ABCD c SA lng cao v y l hnh ch nht ABCD,bit SA = a, AB = b, AD = c.

a) Trong mt phng (SBD), v qua trng tm G ca tam gic SBD mt ng thng ct cnh SB ti M v ctcnh SD ti N. Mt phng (AMN) ct cnh SC ca hnh chp S.ABCD ti K. Xc nh v tr ca M trn cnhSB sao cho th tch ca hnh chp S.AMKN t gi tr ln nht, nh nht. Tnh cc gi tr theo a, b, c.

b) Trong mt phng (ABD), trn tia At l phn gic trong ca gc BAD ta chn mt im E sao cho gc BEDbng 450

( ) ( )2 22 b c 2 b cAE

2

+ + +=. Cmr:

55. Cho hnh chp S.ABCD,y l hnh bnh hnh tm O. Hai mt bn SAB v SCD vung gc ti A v C cng hp viy gc . Bit ABC = . Chng minh SBC v SAD cng hp vi y ABCD mt gc tha mn h thc :

cotcot os.c = .56.Cho hnh chp S.ABC,y ABC l tam gic vung ti B vi AB=a, SA vung gc vi mt phng (ABC) ; mt (SAC)

hp vi mt phng (SAB) mt gc v hp vi mt phng (SBC) mt gc . Chng minh rng :

acos

cos[ ( )].SA

cos( )

+=

57.Cho hnh chp S.ABCD c y ABCD l hnh ch nht ; SA vu ng gc vi mt phng


32


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PHN VI : MT S KIM TRA I TUYN

S GD&T NGH ANTRNG THPT NG THC HA

Gio vin r a : Phm Kim Chung

BI KIM TRA CHT LNG I TUYN THAM GIA K THI HSG TNHNM HC 2010 2011

( Ln th 1 ) Thi gian lm bi : 180 pht_____________________________________

( )2(x 21)l x xn 1 2x

+++ =Cu 1. Gii phng trnh :

Cu 2. Xc nh tt c cc gi tr ca tham s m h phng trnh sau c nghim duy nht :2

2

22

m2x

y

m2y

y

xx

= +

= +

a ,b, c 0>Cu 3. Cho . Tm gi tr nh nht ca biu thc :4a b 3c 8c

Pa b 2c 2a b c a b 3c

+= +

+ + + + + +

( )nx ,n N *Cu 4. Cho dy s , c xc nh nh sau : 12

x3

= v nn 1

n

xx ,

2(2n 1n

)xN *

1+ +

=+

. t

n 1 2 nx ..y xx .+ + += . Tm

nnlim y+

.

Cu 5. Cho hnh chp S.ABCD c SA l ng cao v y l hnh ch nht ABCD, bit SA = a, AB = bAD = c. Trong mt phng (SBD), v qua trng tm G ca tam gic SBD mt ng thng ct cnh SBti M v ct cnh SD ti N. Mt phng (AMN) ct cnh SC ca hnh chp S.ABCD ti K. Xc nh v tr

ca M trn cnh SB sao cho th tch ca hnh chp S.AMKN t gi tr ln nht, nh nht. Tnh ccgi tr theo a, b, c.

1 1 1 1BAB .A CD DCCu 6. Cho hnh lp phng c di bng 1 . Ly im

1E AA sao cho

1AE

3= . Ly

im F BC sao cho1

BF4

= . Tm khong cch t1

B n mt phng FEO ( O l tm ca hnh lp

phng ).

( ) ( )0: 0; ;f + +Cu 7. Tm hm s tho mn :

( ) ( )xf xf(y ) f f( y) x, 0, y ; )( += __________________________Ht__________________________

Thanh Chng ,ngy 03 thng 12 nm 2010


33


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HNG DN GII V P S

( )2(x 21)

l x xn 1 2x+

++ =Cu 1. Gii phng trnh : (1)

x 1> Li gii: iu kin :2 22(x 1)ln(x 1) x 2(x 1)ln(2x 21 0x xx ) + + = + + + =Lc : PT

( ) 2f(x) 2 x 1 ln( 2x, x 1x 1) x= + + > Xt hm s :

f ' (x) 2ln(x 1) 2x= + Ta c : ;

2 2xf ''(x) 2

x 1 x 1

= =

+ +;

2

2f '''(x) 0,

(x 1)x 1=

f ''(0) 0, f '''(0) 0= Do :

( ) 2f(x) 2 x 1 ln(x 1) 2xx= + + Vy hm s nghch bin trn khong ( )1; + . Nhn thy x 0= l mt nghim ca

phng trnh (1), suy ra phng trnh c nghim duy nht x 0= .

Cu 2. Xc nh tt c cc gi tr ca tham s m h phng trnh sau :

22

22

m2x

y

m2y

y

xx

= +

= +

c nghim duy nht .

0;y 0x = =/ /L

i gii : iu kin :2 2 2

2 2 2

2x y

2y

y m

x x m

= +

= +

H cho tng ng vi : (*)

x 0;y 0> >T h (*) nhn thy v tri ca cc phng trnh khng m, nn nu h c nghim (x,y) th :

2 2 2

3 2 2

x 0,y 0y x 0

(*) 2x y2x (1)

(x y)(2xy x )

y

y 0

mx m

> >= >

= =

+ + =

Do :

Do bi ton tr thnh tm tham s m phng trnh (1) c nghim dng duy nht. 3 2

f(x) 2x x , x 0 >=Xt hm s :

2

x 0

f '(x ) 6x 2x; f '(x ) 0 1x3

=

= = =

Ta c :


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2

m 0Nhn vo bng bin thin ta thy, phng trnh (1) c nghim dng duy nht khi v ch khi : . Vy vi mim R h phng trnh cho c nghim duy nht.

a,b,c 0>Cu 3. Cho . Tm gi tr nh nht ca biu thc :

4a b 3c 8cP

a b 2c 2a b c a b 3c

+= +

+ + + + + +

Li gii :

x a b 2c a y z 2x

y 2a b c b 5x y 3z(x,y,z 0)

z a b 3c c z x

= + + = +

= + + = > = + + =

t :

( )4 y z 2x 2x y 8(z x) 4y 2x 4z 8xP 17

x y z x y x z

+ = + = + + +

Lc :

2 8 2 32 17+ 12 2 17=

( )

( )

4 3 2a t

22y 10 7 2

b t t2

2xR,t 0

2x2z 2c 2 1 t

+=

=

= =

=

>

Du = xy ra khi v ch khi :

( )nx ,n N *Cu 4. Cho dy s c xc nh nh sau : 12

x3

= v nn 1n

xx ,

2(2n 1n

)xN *

1+ +

=+

. t

n 1 2 nx ..y xx .+ + += . Tm

nnlim y+

Li gii :

n

n 1

n n 1 n

x 1 1x 2(2n 1)

2(2n 1)x 1 x x+

+

= = + ++ +

T : . t : nn

1v

u= , ta c : 1

n 1 n

3v

2

v 2(2n 1) v+

=

= + +

n 1

(2n 1)(2n 3)v

2+

+ +=D dng tm c cng thc tng qut ca dy :

n 1

n 1

1 1 1 1 1x

v 2n 1 2n 3 2n 1 2(n 1) 1+

+

= = + + + + +

=Do : suy ra :

n 1 2 n 1

1 1 1 1 1 1 1y x x ... 1

2 1 2.2 1 2.2 1 2.3 1 2(n 1) 1 2n 1 2n 1x ... x

= = + + + + = + + + + + + + +

+

+ +

nn n

1lim y lim 1 1

2n 1+ + = = +

Do :

Cu 5. Cho hnh chp S.ABCD c SA lng cao v y l hnh ch nht ABCD bit SA = a, AB = b, AD = c.Trong mt phng (SBD) v qua trng tm G ca tam gic SBD mt ng thng ct cnh SB ti M v ct cnhSD ti N. Mt phng (AMN) ctcnh SC ca hnh chp S.ABCD ti K. Xc nh v tr ca M trn cnh SB sao choth tch ca hnh chp S.AMKN t gi tr ln nht nh nht. Tnh cc gi tr theo a, b, c.

Li gii : Do G l trng tm tam gic SDB, suy ra G cng l trng tm tam gic SAC. Do AG ct SC ti trung im K ca SC.


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1S 1x ; y

2

M SNx, y 1 1

SB 2SD=

=t :

SANK SAKM

SADC SACB

V VSA SN SK y SA SK SM x. . ; . .

V SA SD SC 2 V SA SC SB 2= = = =Theo cng thc tnh t s th tch ta c : Li c

SADC SACD SABC

1 1V V abc

2V

6= == v :

SANK SAKM SANKMV VV + = . Nn ta c :

SANK SAKM SANKM

SANKM

SADC SACB SABCD

V V 2V x y abc(x y)V

V V V 2 12

+ ++ = = = (*)

SM 2SN SD ySD; SM SB xSB; SG SO

S

S

SD B 3

N= = = ==

Ta li c :

1 1SO SD SB SG SN SM

3y 3x2 = + = +

V O l trung im ca BD nn : (1)

1 1 y 11 x

3y 3x 3yy

21

1

+ = =

M : M, N, G thng hng nn t (1) ta c :

2

SANKM

yabc y

3y 1 abc yV

24 8 3y 1

+ = =

Thay vo (*) suy ra :

2y 1f(y) y 1

3y 1 2

=

Xthm s :

( )

2

2

3y 2y 2f '(y) ; f '(y) 0 y

33y 1

= = =

Ta c : .

Bng bin thin :

1y4 2 1

Minf(y) y ; Maxf(y) 29 3 2

y 1

== = =

=

Nhn vo bng bin thin ta thy :

( )SANKMabc

Max V MN / /BD9

=

T ta c :

( )SANKM abcMin V8

= M l trung im SB, hoc N l trung im SD.

1 1 1 1BAB .A CD DCCu 6. Cho hnh lp phng c di bng 1 . Ly im

1E AA sao cho

1AE

3= . Ly im

F BC sao cho1

BF4

= . Tm khong cch t1

B n mt phng FEO ( O l tm ca hnh lp phng ).

1I A(0;0;0);A (0;0;1);D(1;0;0);B(0;1;0)Li gii : Chn h trc ta Ixyz sao cho


36


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1 1 1O ; ;

2 2 2

Lc : O l trung im AC nn ; ( )11 1

E 0;0; ;F ;1;0 ;B 0;1;13 4

1 5 3; ;

3 24OE,OF

8

=

Mt phng (OEF) i qua O v nhn vct lm vct php tuyn nn c ph ng trnh :

1 1 5 1 3 1x y z 0

3 2 24 2 8 2

=

hay : 8x 5y 9z 3 0 + =

( )12 2 2

5 9 3 11d B (OEF)

1708

;

5 9+

=

+

+=Vy :

( ) ( )0: 0; ;f + +Cu 7. Tm hm s tho mn : ( ) ( )xf xf( y) f f( y) x, 0, y ; )( +=

Li gii :

( ) ( )xf xf(1) f f(1)=Cho y = 1, suy ra : . t f(1) a= , ta c : xf(ax) f(a)= (1)

1x

a=T (1) cho , suy ra :

1f(1)=f(a) f(a)=1

a

1f(ax)

x=Cng t (1) cho ta : (2)

aax y f(y)

y= =T (2) cho

af(y) (a 0)

y= >Th li ta thy l hm s cn tm .


37


38/46

S GD&T NGH AN TRNG THPT NG THC HA

Gio vin ra : Phm Kim Chung (S&GT) - Nguyn Th Tha (HH)

BI KIM TRA CHT LNG I TUYN THAM GIA K THI HSG TNHNM HC 2010 2011

( Ln th 2 )Thi gian lm bi : 180 pht

_____________________________________

3 23x 3 5 2x x 3x 10x 26 0+ + + =Cu 1. Gii phng trnh:

Cu 2. Tm tt c cc gi tr ca tham s m h phng trnh sau c nghim:2 2

3 3

2 2

3 3

log y

log

x lo

y

g 1 2m 3

log x 1 2m 3

+ =

+ =

a, b, cCu 3. Cho dng tho mn ab bc ca abc+ + = . Chng minh rng:

2 2 2 2 2 2

b c a 1 1 1

a b c a c3

b

+ + + +

4 3 2f(x) x ax bx cx d+ + + +=Cu 4. Cho hm s . Ta k hiu o hm bc n ( n nguyn dng ) caf(x) l ( )(n )f x . Chng minh rng nu f(x) > 0, x R th :

(1) (2) (3) (4)F(x) (x) f f (x) f (x) 0,( x Rf x) f + + + > + =

DACu 5. Cho t din ABCD c vung gc vi mt phng(ABC), tam gic DAB cn v y ABC ltam gic vung ti B c BAC = . Gi l gc to bi hai mt phng ( )DAC v ( )DBC . Chng minh

rng:21 cos

tan .tancos

+ =

.

ABCD.A'B'C'D'Cu 6. Cho hnh lp phng cnh a . Vi M l mt im thuc cnh AB, chn imN thuc cnh D'C' sao cho AM D'N a+ = . Tnh th tch khi chp B'.A'MCN theo a v xc nh v tr

ca im M khong cch t im B' n mt phng ( )A'MCN t gi tr ln nht.

f :R RCu 7. Cho hm s tha mn h iu kin :

4

x, y R

f(1) 1

f (x y ) f (x) f (y ) 2xy ,

1 f(x)0f ,

x xx

=

+ = + + =

=/

.

( )

32f(x)

x 0

1 f1

e(x

L limln 1 f )

)

(x

+

+=Tnh gii hn :

---------------------------------Ht------------------------------

Thanh Chng, ngy 10 thng 12 nm 2010


38


39/46

HNG DN GII V P S

3 23x 3 5 2x x 3x 10x 26 0+ + + =Cu 1 . Gii phng trnh :

21

5x Li gii:K:

( ) ( ) 2

2

2

3x 3 3 5 2x 1

3x 3 3 5 2x 1

x 12)3x 3 3 5 2x

PT (x 2)(x x 12) 0

3(x 2) 2(x 2) (x 2)(x x 12) 0

3 2( (

1x 2) x 0

+

+ + +

=

+ =

+ =

+ + +

2 x 12,5

f(x) x 1;2

x

= +

+

Xt hm s :

1f '(x) 2x 1, f '(x) 0 x

2= + = =Ta c: . Suy ra:

51

2;

1Minf(x) Min f( 1);f ;f f 0

2

5 5

2 2

= = >

23 2 5(x x 12) 0, x 1;23x 3 3 5 2x 1

+ >

+ + + Do :

Vy phng trnh c nghim duy nht: x = 2 .

2 2

3 3

2 2

3 3

log y

log

x lo

y

g 1 2m 3

log x 1 2m 3

+ =

+ =

Cu 2 . Tm tt c cc gi tr ca tham s m h phng trnh sau c nghim :

x,y 0>Li gii: K:

( )2

3

2

3

uu

v y

log x 11, v 1

log 1

+=

+

=t: . Lc h PT tr thnh :

2

2

u v 2m 2 (1

u 2m 2 (2)

)

v =

=

.

Ly (1)-(2), ta c : ( )( )u v u v 1 0 u v + + = = ( Do u+v+1 > 0 u, v 1 )

Lc bi ton tr thnh tm m phng trnh : 2 u m 2u 2 = c nghim u 1 .

Xt hm s : 2f(u) u u 2= + , ta c : f '(u) 2 uu 1 , 10= > . Vulim f(u) +

= + .

Do , PT trn c nghim u 1 khi v ch khi f(12m u) 2 1f( ) m= =

ab bc ca abc1 1 1

1a b c

++ = ++ =Li gii: Ta c: . t: ( )1 1 1

x y z 1 x,y,x, y,a c

zb

0z + + == = >= . Bt ng thc

cn chng minh tr thnh: ( )2 2 2

2 2 2x y z xy z x

3 y z ++ ++ .

4 4 4 2 2 2 2

2 2 2 2 2 2

x y z ( y z )

y

xVT

x y z xz x y y z z x

+ +

+ += + +p dng BT Svac-x ta c: .

( )2 2 22 2 22 2 2 2 2 2

(x y yz) xx

x

zy z

x3 3

y y z z x y y z z x

+ + ++ +

+ + + +

+Ta s chng minh: ( do x y z 1+ + = )

3 2 3 2 2 3 2 2 2xy y yz zxx 2(x y yz z z x) + + + + + + + (*)

3 2 4 2 2 3 2 4 2 3 2 4 2xy 2 x y 2x y; y yz 2 y z ; z z 2 z xx x+ = + + Theo bt ng thc AM -GM ta c : . Cng cc BT trn

ta chng minh c (*). Vy: 2 2 2V (xT 3 y z ) + + . pcm

a, b, cCu 3. Cho dng tho mn ab bc ca abc+ + = . Chng minh rng :2 2 2 2 2 2

b c a 1 1 1

a b c a c3

b

+ + + +


39


40/46

4 3 2f(x) x ax bx cx d+ + + +=Cu 4. Cho hm s . Ta k hiu o hm bc n ( n nguyn dng ) ca f(x) l ( )(n )f x .

Chng minh rng nu f(x) > 0, x R th : (1) (2) (3) (4)F(x) (x) f f (x) f (x) 0,( x Rf x) f + + + > + =

xlim F(x)

= +Li gii : Ta c: F(x) l hm bc 4 v : , hn na phng trnh bc 3 : F(x)=0 lu n c nghim. Do

hm s y = F(x) lun c GTNN l gi tr cc tiu ca hm s .

0x x=Gi s hm s t cc tiu ti .

0F '( x ) 0=Lc : suy ra: 1 2 3 40 0 0 0 0 0 0) f (x ) f (x ) f (x ) f (x ) F(x0 F' ) f )( (xx = + = + + =

0 0) f(x ) 0F(x = > ( Do f(x) > 0 , x R )

0F(x ) 0 xF(x R) , > T ta c:

DACu 5 . Cho t din ABCD c vung gc vi mt phng (ABC),tam gic DAB cn v y ABC l tam gic vung ti

B c BAC = . Gi l gc to bi hai mt phng ( )DAC v ( )DBC . Chng minh rng :21 cos

tan .tancos

+ =

.

Li gii : t DA = x. Gi K l hnh chiu ca A ln DB, t K kKH vung gc vi DC ti H . Ta c :

BC DBBC (D

DA BC

AAB)

B B BC AC K

. Suy ra :

AK (DBC)AK KH

DC AK

DC (AHK) AH DC

AHK = Do :AK

tanHK

= (1)

BCBC A B. tan x. t t

Aa n

Bn a = = = Trong tam gic ABC : ;

AB xAC

Ao

oCc

cs

s = =

;

2 2 2

1 1 1

AK AD AB

x 2AK

2 == + Trong tam gic ADB : (2) : 2 2AD AB 2BD x+ ==

2 22

2 2 2 2 2 2

1 1 1 1 cos xAH

AH AD AC x x 1 cos= = + =

+

+

Trong tam gic ADC:2

2 2 2 2

2 2

x xcosDH x DH

1 cos 1 cosAD AH

= =

+ =

+

( )2

x tan .xcosDH DB BC.DHDHK ~ DBC HK

HK BC BD 1 c . xos 2

= = =

+ Xt hai tam gic vung: (3)

2 2

2

AK x 2 2x. 1 c 1 ctan . tan .tan

HK 2 cx .ta

os os

o. sn osc

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Cac Bai Toan Boi Duong Hsg 2010