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8/8/2019 [Www.vnmath.com]-De Thi Lop 10 PTNK TpHCM 2009- 2010
1/13
I HC QUC GIA TP HCM K THI TUYN SINH L P 10 NM HC 2009 - 2010TR NG PHTHNG Mn thi: TON
NNG KHIU Th i gian lm bi: 120 pht, khng k th i gian giao ___________________________________________________________________________________ ___ _
Bi 1. (2im)
a) Gii ph ng trnh bng ccht n s 54 x
t x =
:
22400 535 24 4
x x x x
+ = +
b) Cho ph ng trnh ( )2 3 1 2 3 0mx m x m+ + + = .Tm m ph ng trnh c hai nghim phn bit 1 2, x x tha mn 2 21 2 34 x x+ =
Bi 2. (2.5im) Xt biu thc: 2 3 3 4 51 5 4 5
x x x x R
x x x x
+ + + = +
a) Rt gn R . b) Tm sthc x 2 R > . Tm stnhin x l schnh ph ng sao cho R l snguyn.
Bi 3. (2im)
a) Gii hph ng trnh: 2 20
8 x xy y
x y
+ + =
+ =
b) Cho , ,a b c l di ba cnh ca tam gic ABC . Gisph ng trnh( )( ) ( )( ) ( )( )0 x a x b x b x c x c x a + + =
c nghim kp. Tnh s o cc gc ca tam gic ABC .
Bi 4. (1.5im)Cho tam gic ABC , c
0 060 , 45 ABC ACB= = . Dng ( ) AH BC H BC , v dng( ) HK AB K AB . Gi l trungim ca AC . Bit 3 AH = , tnh BC . Chng minh
BKMC l tgic ni ti p.
Bi 5. (1im)Trong k kim tra mn Ton mt l p gm 3 tA, B, C,im trung bnh ca hc sinh cc t c thng k bng sau:
T A B C A v B B v Cim trung bnh 9.0 8.8 7.8 8.9 8.2
Bit tA gm 10 hc sinh, hy xcnh shc sinh vim trung bnh ca ton l p.
Bi 6. (1im)Cho tgic li ABCD ni ti p ng trn( )O , cnh A c nh v ccnh , , B C D dichuyn trn( )O sao cho 090 BAD > . K tia Ax vung gc v i AD ct BC ti E , k tia Ay vung gc v i AB ct CD ti F . Gi K l im i xng ca A qua EF . Chng minh t gic EFCK ni ti p c v ng thng EF luni qua mt im c nh.
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H ng dn giiBi 1.
a) t 54 x
t x =
, suy ra2
2 2 22 2
5 25 400 5162 16 2
xt x t
x x
+ = + + = +
Ph ng trnh tr thnh 2 416 24 5 01
4
t t t
t
5 =
+ = =
V i 54
t = , ta c 1,25 5 5 105
4 4 2 x
x x
= =
V i 14
t = , ta c 34
55 144 4
x x x x
= = =
Vy 5 105 5 1055;4; ;2 2
S + =
b) iu kin ph ng trnh c hai nghim phn bit( ) ( )2
00
9 1 4 2 3 0m
mm m m
= + + >
V i iu kin trn, ph ng trnh c hai nghim phn bit x1, x2 v theonh l Viet ta c( )
1 2
1 2
3 1
2 3
mS x x
mm
P x xm
+= + =
+ = =
Khi ( )( )
( )
222 2
1 2 1 2 1 2 2
113 12 934 2 34 34 37
m nm m
x x x x x xm m n
=+ + + = + = =
=
p s: 31, 7m m= = Bi 2
a) t t x= ta c( )( ) ( )( )( )
( )( )
( )( )( )( )( )( )
22
2
2
2 5 3 1 3 4 52 3 3 4 51 5 1 54 5
1 23 2 2 21 5 1 5 5 5
t t t t t t t t t t R
t t t t t t
t t t t t xt t t t t x
+ + + + + + + + = =+ +
+ + + += = = = + +
b)* iu kin 0
Ta c 52 2 122 2 2 0 0125 5 5
t t t t Rt t t t
> > > >
V i 5 5 0 25t x x< < < V i 12 12 144t x x> > > Vy gi tr cn tm l0 25 < v 144 x > Ta c l schnh ph ng nnt x=
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Khi 2 715 5
t R
t t
+= = +
t 5 l c ca 7, mt khc 5 5t do
5 1,1,7t = T nhng gi tr x cn tm l 16,36,144 x =
Bi 3.
a) 2 20
8 x xy y
x y
+ + =
+ =
t ,S x y P xy= + = , khi ta c h 2 2
40 4
22 8 2 8 02
S S P P S P
S S P S S
P
= + = = =
= = + = =
V i4
4S
P
= =
ta c4 2
4 2 x y x
xy y
+ = = = =
V i2
2S
P
= =
ta c2
2 x y
xy
+ = =
gii hta c1 31 3
x
y
=
= +hoc
1 31 3
x
y
=
= +
Vy hph ng trnh c 3 nghim ( ); y l ( ) ( ) ( )2; 2 , 1 3;1 3 , 1 3,1 3 + + b) ( )( ) ( )( ) ( )( )0 x a x b x b x c x c x a + + =
( ) ( )23 2 0 x a b c x ab bc ac + + + + + = Ta c ( ) ( )2 2 2 23a b c ab ac bc a b c ab ac bc = + + + + = + + Ph ng trnh c nghim kp khi v ch khi
( ) ( ) ( )2 2 22 2 2 10 0 020
a b c ab bc ac a b b c c a
a b b c c a a b c
= + + = + + =
= = = = =
Khi tam gic ABCu, suy ra 0
60 A B C = = = Bi 4.
a) Trong tam gic vung ABH ta c
03tan 1
tan60tan AH AH
ABH BH BH ABH
= = = =
Trong tam gic vung AHC c0 045 45 ACH HAC = = nn AHC l tam gic vung
cn, suy ra 3 HC HA= = Do 1 3 BC BH CH = + = + (vd) b) Tam gic AHC vung cn, c AM l trung tuyn nncng l ng cao, suy ra AM HC
C1: Tgic AKHM c 0 0 090 90 180 AKH AMH + = + = nn l tgic ni ti p, suy ra0 090 45 AKM AHM HAM = = =
Tgic BKMC c 045 AKM BCM = = nn l tgiv ni ti p.C2: Ta c AK. AB = AH2, AM. AC = AH2, suy ra AK. AB = AM. AC
45 0 60 0
K
H CB
A
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Suy ra tam gic AKM v ABCng dng (c.g.c), suy ra 045 AKM BCM = = nn l tBKMC gicni ti p.
Bi 5.Gi x, y ln l t l shc sinh tB v C.
Ta c 9 10 8,8 8,9 1010
x x
x
+ = =+
T ng t 8,8 7,8 8,2 x y x y
+ =+
, v i x = 10 th y = 15
Vy im trung bnh ca cl p l 9 10 8,8 10 7,8 15 8,4310 x y
+ + =+ +
Bi 6.* Tgic ABCD ni ti p nn 0180 BAD BCD+ = V
0 0 090 90 180 BAD EAF BAE EAF FAD EAF
BAF DAE
+ = + + +
= + = + =
Suy ra BCD EAF = (1)Mt khc, do A v K i xng nhau qua EF nn EKF EAF = (2)T(1) v (2) suy ra
EKF ECF = , do tgicEFKC ni ti p.* V tgic EFKC ni ti p nn ta c FCK FEK = m FEK FEA= (do tnh cht i xng)V FEA KAD= (cng phv i KAE )Do KAD FCK =
Suy ra tgic ADKC ni ti p, suy ra K thuc (O), suy ra OA = OK, suy ra O thuc ng trung tr cca AK m EF l ng trung tr c ca AK nn O thuc EF. Vy ng thng EF luni quaim Oc nh.
Nhn xt:
nm nay cho kh di so v i th i gian 120 pht. V lchung cho tt ccc l p chuyn nn kinthc dn tr i v c vi cu kh. Tuy nhin c 4 th cng khng kh nu cc em lm bi cn thn.C thnhn xt tng cu nhsau:Cu 1:
a) (0,75) Cu ny nhiu em khng lm c, v khng thtnh tt ctheot . b) (1,25) Cu ny thuc dng c bn v d, cc em stiu kin ph ng trnh c hai nghim phn
bit. (0,25) v nhiu em khng hiu sao li btr ng h p m = -3/7 (!)Cu 2. a) (1) Cu ny dv quen, v quan tr ng v nu lm c th cu b m i lm c. Tuy vy cnhiu khng rt gn trit hoc sai du (!)
b) *(1) Cu ny nhiu bn sai nht, v khng chuyn vxt tr ng m quyng bmu mt cchr t tnhin v tt nhin l sai.(*) (0,5) Cu ny khng kh v nhiu em lmng.
Cu 3. a) Bi hth qu c bn, tuy vy c nhiu em gii ra tch v tngng nhng khi p dngnhl o Viet li sai (X2 SX + P = 0 m c ln X2 + SX P = 0)
y
x
K
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O
A
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b) Bi ny nhn c vr c r i nhng nu a vph ng trnh bc hai th coi nhxong. (li mt cu v ph ng trnh bc 2)Cu 4. Cu ny c ll dnht trong, v hu lt lm c vng.Cu 5. Cu ny khng kh, nu chu lm th slmng k t qu. V cng nhiu em lmng.Cu 6. Cu ny l cu kh nht, v nhiu em bnht. u tin c lkhng kh nhng sau th k h. Cu ny l cu phn loi v dnh cho hc sinh chuyn ton.
Trny l mt vi nhn xt chquan ca ng i vit. Hy vng rt kinh nghim trong cc k thi sau vc k t qutt h n.
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I HC QUC GIA TP HCM K THI TUYN SINH L P 10 NM HC 2009 - 2010TR NG PHTHNG Mn thi: TON CHUYN
NNG KHIU Th i gian lm bi: 150 pht, khng k th i gian giao _______________________________________________________________________________
Cu 1.
a) Cho , , ,a b c d l cc s thc tha mn iu kin , . 03a c a c
a cb d b d
+= = .
Chng minh r ng: 2 2b d = . b) Gi i hph ng trnh:
2 2
2 2
1 33 7
2 34 7
x x y xy x y
y x y xy x y
= =
Cu 2.a) Gii bt ph ng trnh: 2 1 8 9 x x+ +
b) Cho , ,a b c l cc s thuc [ ]1;2 tha mn iu kin 2 2 2 6a b c+ + = .Chng minh r ng: 0a b c+ +
Cu 3.a) Chng minh r ng khng t n ti stnhin a sao cho
2 20092010a a+ = b) Ch ng minh r ng khng t n ti stnhin a sao cho
2 3 20102009a a a+ + =
Cu 4.Cho ng trn ( )O tm O , ng knh 2 AB R= . C l mt im thay i trn ng trn( )O sao cho tam gic ABC khng cn t i C . Gi H l chn ng cao c a tam gic ABC ht C . H , HE HF vung gc v i , AC BC t ng ng. Cc ng thng EF v AB ctnhau t i K .
a) Tnh theo R din tch tam gic CEF v di cc on ,KA KB trong tr ng h p060 BAC = .
b) H , EP FQ vung gc v i AB . Chng minh r ng ng trn ng knh PQ ti pxc v i ng thng EF .
c) Gi D l giao im ca ( )O v ng trn ng knh CH , D C . Chng minhr ng 2.KA KB KH = v giao im M ca cc ng thng CD v EF lun thu c mt ngthng c nh.
Cu 5.Trn m t ng trn, ng i ta x p cc s 1,2,3,...,10 (m i sxut hin ng m t ln).
a) Chng minh khng t n ti mt cch x p m t ng hai s k nhau u l n h n 10. b) Tn ti hay khng m t cch x p m t ng hai s k nhau u l n h n hoc bng 10?
-----------Ht------------
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H ng dn gii D i y ch l h ng d n gi i chquan c a chng ti v khng ph i l p n
chnh th c ca tr ng nn mang gi tr tham kh o l chnh.Bi 1.
a)Tr ng h p 1: 2 2b d b d = = (ccm)
Tr ng h p 2: b d , k t h p v i iu kin a cb d
= suy ra a c
Khi a c a cb d b d
+= =+ (tnh ch
t dy t sbng nhau)
Suy ra0
33
a ca c a cb d b d b d b d
+ =+ += = + +
V i 0a c+ = m 0ac suy ra 0, 0a c suy ra b d = (mu thu n)V i 2 23b d b d b d b d = + = = Vy trong hai tr ng h p ta u c 2 2b d =
Nhn xt:My em p ng ngay dy t sbng nhau l thi u tr ng h p r i, sb tr im. b)
2 2
2 2
2 2
1 33 7 1 2 3
2 3 3 4 74 7
x x y xy x y x y x y
y x y xy xy x y xy x y
=
= = =
iu kin2 2
3
4
7
xy
xy
x y
+
Tr ng h p 1: ( ) 73 42
xy xy xy = = , khi ( )1 2 3 x y x y = + =
Ta c h ( )3
72
x y
VN xy
+ =
=
Tr ng h p 2: ( )3 4 xy xy
Khi ta c 2 23 1 2 3
7 3 4 2 7 x y x y x y
x y xy xy xy
+ = = =
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Suy ra( )2 2
3 0
7 2 7
x y
x y xy
= =
V i 3 0 x y = ta c1
1 2 02
x x y
y
= = = =
V i ( )22 2
7 2 7 0 x y xy x y x y = + = =
Khi ta c 2 21 11 2
2 23 4
x y x x x y x x
= = = = =
Th li ta thy ( )1;2 v ( )1; 1 l nghi m ca hph ng trnhVy ph ng trnh c hai nghi m ( ); x y l ( )1;2 v ( )1; 1 Nhn xt: Bi h ph ng trnh t ng c ng ging cu a, dng dy t sbng nhau. Khng
kh, tuy nhin l i dsai, v thi u st. V d 0mm m
x y x y
== =
(dst tr ng h p 0m = )
Bi 2a) Ta c
( )
( )( )2
2 1 0
8 9 02 1 8 9
2 1 0
2 1 8 9
x I
x x x
x II
x x
+
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b) V [ ] ( )( ) 2 21;2 1 2 0 2 0 2a a a a a a a + Du = x y ra khi v ch khi 1a = hoc 2a =
Chng minh t ng t ta cng c 2 22, 2b b c c
Do 2 2 2 6a b c a b c+ + + + suy ra 0a b c+ + (v 2 2 2 6a b c+ + = )
Du = x y ra khi v ch khi ( ), ,a b c l hon v ca ( )1; 1; 2 Nhn xt:y l bi ton b t ng thc c iu kin khng n gin cht no so v i l i giica n. hgc nhi u em. Tuy nhin n u ch th cu a g i t ng lm cu ny.Cu 3a) Gis tn ti stnhin a tha 2 20092010a a+ =
Ta c ( )2 1a a a a+ = + l tch hai s tnhin lin ti p.Ta c ( ), 1 1a a + = v ( )1 1a a+ = .
Do , 1a a + phi c dng 2009 2009, 1a p a q= + = trong p q< ,( )
. 2010, , 1 p q p q= =
iu ny khng th xy ra v ( ) ( )20092009 2009, 1 1 1 1 p q q p q p p= + > + Vy khng t n ti stnhin a tha mn bi.Nhn xt:Bi ny hi u nh ng kh trnh by qu, d r i vo tnh tr ng lng vng. Kinhnghim th khi cho s l n th ng khng nh h ng n cch gi i, tuy nhin i v i bi ny vsm l lnn khng th dng tnh ch t ca schnh ph ng c. H n na, khng th xttheo modul 3, 4 v n th a ht. Ci hay l cu a v b nhn c v ging nhau nh ng cch gi i li
khc nhau.
b) Gi s tn ti stnhin a tha bi. Tc l 3 2 20102009a a a+ + = R rng 0a > , khi ta c ( )33 3 2 3 23 3 1 1a a a a a a a a< + + < + + + = +
Mt khc ( )32010 6072009 2009=
Suy ra ( ) ( )3 33 6702009 1a a< < + . (V l v ( )33 , 1a a + l l p ph ng ca hai s tnhin lin
ti p. )
Vy khng t n ti stnhin a tha mn bi.Nhn xt
Bi ny thu c dng quen thu c ca ph ng trnh nghi m nguyn, nh ng i khi b nhiu b i cu a, kh nh n ra.
Ni chung n m nay hai bi s hc khng kh b ng bi s hc nm ngoi (Bi v s bchkim )Cu 4
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J
T
D
I
P QK,M
F
E
H OA B
C
a) Tnh theo R din tch tam gicCEF v di ccon ,KA KB trong tr ng h p
060 BAC = .
Ta c 090 ACB = (gc n i ti p chn na ng trn ( )O )
Tam gic ABC vung t i C nn ta c 0.cos 2 .cos60 AC AB CAB R R= = =
V 0.sin 2 .sin 60 3CB AB CAB R R= = =
Ta c 03
.sin .sin 602
RCH AC ACB R= = =
Tam gic CHE vung t i H c HE l ng cao nn2
22
3
2 3.4
R
CH CE CA CH CE RCA R
= = = =
T ng t ta cng c2 3
4CH R
CF CB
= =
Do 21 1 3 3 3 3
. . .2 2 4 4 32CEF
R R RS CE CF = = =
V 060 BAC = nn A nm gia K v B
Dthy CEHF l hnh ch nht v 030KEA CEF CHF CBA= = = = , m0 0 060 30 30 AKE AEK CAB AKE CAB AEK + = = = =
Vy tam gic KAE cn ti A suy ra KA AE =
M3 14 4 R
AE AC CE R R= = = nn 14
KA R=
V1 9
24 2
KB KA AB R R R= + = + =
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b) Ch ng minh EF tip xc v i ng trn ng knh PQ
Cu b, c ta xt tr ng h p AC < BC, tr ng h p AC BC > lm t ng t Gi I l giao im ca EF v CH . V AEHF l hnh ch nht nn I l trung im EF .Tgic EPQF l hnh thang vung (v , EP FQ PQ )
Ta c // IH EP v I l trung im EF nn H l trung im ca PQ .Khi ng trn ng knh PQ l ng trn tm H bn knh HP .Gi T l hnh chi u ca H trn EF
Ta c
PEH EAH = (cng ph EHA ) v TEH IHE = , IHE EAH = (cng ph v i EHA . )
Suy ra PEH TEH = , suy ra PEH TEH HT HP = = Ta c ( ) HT EF T EF v HT HP= nn EF ti p xc v i ng trn ng knh PQ
c) Ch ng minh 2.KA KB KH = v M thuc mt ng c nh
Ta c KEA CEF CHF CBK = = = , suy ra ( ).KAE KFB g g ,
Do . .KA KE KA KB KE KF KF KB
= = (1)
Mt khc ta c
KHE HCE HFK = = , suy ra ( ).KHE KFH g g
Do 2.KH KE KE KF KH KF KH
= = (2)
T (1) v (2) th 2.KA KB KH =
Gi J l giao im ca OC v EF,
Ta c OCF OBC = (tam gic OBC cn t i O)
V
JFE ICF = (do tam gic ICF cn t i I)Do
0
0
90
90
OCF JFE OBC ICF
CJF OC EF
+ = + =
=
Tam gic CKO c CH v KJ l hai ng cao, c t nhau t i I nn I l tr c tm c a tam gic
CKO, do OI CK (3)Mt khc hai ng trn (O) v ng trn tm I ng knh CH c t nhau t i C v D, nn OIl ng trung tr c ca CD, suy ra OI CD (4)T (3) v (4) ta c , ,C K D thng hng.
Vy K c ng l giao im ca CD v EF, do M K v M lun thu c ng thng AB c nhNhn xt:y l m t bi hnh h c r t quen thu c, khng kh. h n nm ngoi nhi u.
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Bi 5.a) Gis tn ti mt cch s p x p tha bi l
a 5
a 6 a 7
a 8
a 9
a 10
a 4
a 3
a 2
a 1
Khng m t tnh t ng qut ta gi s 1 1a = . Khi ta c
1 2 2 2
1 10 10 10
10 9 10
10 9 10
a a a a
a a a a
+ > > = + > > =
(v l v m i sxut hin ng m t ln)
Vy khng t n ti cch s p x p tha mn bi. b) Tn ti cch s p x p nh trn. V d :
3
7 4
6
5
9
8
2
10 1
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Nhn xt:Th ng th bi ny nh r t hc sinh ngay t lc c , v b tm l. Nh ngthc sbi ny khng kh b ng nh ng tr c. Khng lm c cu a th c ng li c cu b.
Nhn xt chung v nm nay:nm nay khng kh nh ng cng khng d dng g im cao v c nhi u chby.
Theo ti ngh cu dnht l 2a v 4a. Cu trung bnh l cc cu 1a, 1b, 4b cu kh h ncht l 3a, 3b cc cu kh nh t l 2b, 5ab.
T lchi cao, im chu n cao v Ph Thng N ng Khi u lun ch n c hc sinh gi i.Sang n m c World Cup nn ch c phi c m t cu v bng , hy ch xem.
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