[VNMATH.com] 77 de Thi Vao Cac Lop 10 Truong Chuyen 2013

TUYN TP THI VO LP 10 CC TRNG CHUYN - NNG KHIU, NM HC 2013 - 2014.

Trn Trung Chnh (Su tm).

DANH SCH 77 TRNG IM,

CHUYN, NNG KHIU

TI VIT NAM

STT TN TRNG TNH/

THNH PH

QUN/HUYN/ THNH PH/

TH X

1 Trng Trung hc ph thng Chuyn i hc S phm H Ni

H Ni Cu Giy

2 Trng Trung hc ph thng chuyn Khoa hc T nhin, i hc Quc gia H Ni

H Ni Thanh Xun

3 Trng Trung hc ph thng chuyn ngoi ng, i hc Quc gia H Ni

H Ni Cu Giy

4 Trng Trung hc ph thng chuyn H Ni - Amsterdam H Ni Cu Giy

5 Trng Trung hc ph thng Chu Vn An, H Ni H Ni Ty H

6 Trng Trung hc ph thng Sn Ty H Ni Sn Ty

7 Trng Trung hc ph thng chuyn Nguyn Hu H Ni H ng

8 Trng Ph thng Nng khiu, i hc Quc gia Thnh ph H Ch Minh

Thnh ph H Ch Minh

Qun 10

9 Trng Trung hc thc hnh, i hc S Phm Thnh ph H Ch Minh

Thnh ph H Ch Minh

Qun 5

10 Trng Trung hc ph thng chuyn L Hng Phong, Thnh ph H Ch Minh

Thnh ph H Ch Minh

Qun 5

11 Trng Trung hc ph thng Nguyn Thng Hin, Thnh ph H Ch Minh

Thnh ph H Ch Minh

Tn Bnh

12 Trng Trung hc ph thng Gia nh Thnh ph

H Ch Minh Qun Bnh Thnh

13 Trng Trung hc ph thng chuyn Trn i Ngha Thnh ph

H Ch Minh Qun 1

14 Trng Trung hc ph thng chuyn Thoi Ngc Hu An Giang TP.Long Xuyn

15 Trng Trung hc ph thng chuyn Th Khoa Ngha An Giang TP.Chu c

16 Trng Trung hc ph thng chuyn Trn Ph, Hi Phng Hi Phng Ng Quyn

17 Trng Trung hc ph thng chuyn L Qu n Nng Sn Tr

18 Trng Trung hc ph thng chuyn L T Trng Cn Th Q.Bnh Thy

19 Trng Trung hc ph thng chuyn Nguyn Tt Thnh, Yn Bi

Yn Bi Yn Bi

20 Trng Trung hc ph thng chuyn Thi Bnh Thi Bnh TP Thi Bnh

21 Trng Trung hc ph thng chuyn Lng Vn Ty, Ninh Bnh

Ninh Bnh Ninh Bnh

22 Trng Trung hc ph thng chuyn Vnh Phc Vnh Phc Vnh Yn

www.VNMATH.comwww.VNMATH.com



23 Trng Trung hc ph thng chuyn Bc Giang Bc Giang TP Bc Giang

24 Trng Trung hc ph thng chuyn Bc Kn Bc Kn Bc Kn

25 Trng Trung hc ph thng chuyn Bc Ninh Bc Ninh Bc Ninh

26 Trng Trung hc ph thng chuyn Cao Bng Cao Bng Cao Bng

27 Trng Trung hc ph thng chuyn Nguyn Tri Hi Dng TP Hi Dng

28 Trng Trung hc ph thng chuyn Lo Cai Lo Cai Lo Cai

(thnh ph)

29 Trng Trung hc ph thng chuyn Hong Vn Th Ha Bnh Ha Bnh

(thnh ph)

30 Trng Trung hc ph thng chuyn Tuyn Quang Tuyn Quang Tuyn Quang

(thnh ph)

31 Trng Trung hc ph thng chuyn H Giang H Giang H Giang

(thnh ph)

32 Trng Trung hc ph thng chuyn Chu Vn An Lng Sn Lng Sn

(thnh ph)

33 Trng Trung hc ph thng chuyn L Qu n in Bin in Bin Ph

34 Trng Trung hc ph thng chuyn L Qu n Lai Chu Lai Chu

(th x)

35 Trng Trung hc ph thng chuyn Sn La Sn La Sn La

(thnh ph)

36 Trng Trung hc ph thng chuyn Thi Nguyn Thi Nguyn P.Quang Trung

37 Trng Trung hc ph thng chuyn Hng Vng, Ph Th

Ph Th Vit Tr

38 Trng Trung hc ph thng chuyn L Hng Phong, Nam nh

Nam nh Nam nh

39 Trng Trung hc ph thng chuyn Bin Ha H Nam Ph L

40 Trng Trung hc ph thng chuyn H Long Qung Ninh TP H Long

41 Trng Trung hc ph thng chuyn Hng Yn Hng Yn Hng Yn

42 Trng Trung hc ph thng chuyn Lam Sn, Thanh Ha Thanh Ha Thanh Ha

43 Trng Trung hc ph thng chuyn Phan Bi Chu, Ngh An

Ngh An Vinh

44 Trng Trung hc ph thng chuyn, Trng i hc Vinh, Ngh An

Ngh An Vinh

45 Trng Trung hc ph thng chuyn H Tnh H Tnh H Tnh

46 Trng Trung hc ph thng chuyn Qung Bnh Qung Bnh ng Hi

47 Trng Trung hc ph thng chuyn L Qu n, Qung Tr

Qung Tr ng H

48 Quc Hc Hu Tha Thin-Hu Hu

49 Trng Trung hc ph thng chuyn Bc Qung Nam Qung Nam Hi An

50 Trng Trung hc ph thng chuyn Nguyn Bnh Khim Qung Nam Tam K




51 Trng Trung hc ph thng chuyn L Khit Qung Ngi Qung Ngi (thnh ph)

52 Trng Trung hc ph thng chuyn L Qu n, Bnh nh

Bnh nh Quy Nhn

53 Trng Trung hc ph thng chuyn Lng Vn Chnh Ph Yn Tuy Ha

54 Trng Trung hc ph thng chuyn L Qu n, Khnh Ha

Khnh Ha Nha Trang

55 Trng Trung hc ph thng chuyn L Qu n, Ninh Thun

Ninh Thun Phan Rang -

Thp Chm

56 Trng Trung hc ph thng chuyn Trn Hng o, Bnh Thun

Bnh Thun Phan Thit

57 Trng Trung hc ph thng chuyn Thng Long - Lt Lm ng TP. Lt

58 Trng Trung hc ph thng chuyn Nguyn Du, k Lk k Lk Bun Ma Thut

59 Trng Trung hc ph thng chuyn Hng Vng Gia Lai Pleiku

60 Trng Trung hc ph thng chuyn Nguyn Tt Thnh, Kon Tum

Kon Tum Kon Tum

(thnh ph)

61 Trng Trung hc ph thng chuyn Lng Th Vinh, ng Nai

ng Nai Bin Ha

62 Trng Trung hc ph thng chuyn L Qu n, Vng Tu

B Ra - Vng Tu

Vng Tu

63 Trng Trung hc ph thng chuyn Bn Tre Bn Tre Bn Tre

64 Trng Trung hc Ph thng Chuyn Quang Trung, Bnh Phc

Bnh Phc ng Xoi

65 Trng Trung hc ph thng chuyn Tin Giang Tin Giang M Tho

66 Trng Trung hc ph thng chuyn V Thanh Hu Giang V Thanh

67 Trng Trung hc ph thng chuyn Bc Liu Bc Liu Bc Liu

(thnh ph)

68 Trng Trung hc ph thng chuyn Phan Ngc Hin C Mau C Mau

69 Trng Trung hc ph thng chuyn Hng Vng Bnh Dng Th Du Mt

70 Trng Trung hc ph thng chuyn Hunh Mn t Kin Giang Rch Gi

71 Trng Trung hc ph thng chuyn Nguyn Bnh Khim Vnh Long Vnh Long

72 Trng Trung hc ph thng chuyn Tr Vinh Tr Vinh Tr Vinh

(thnh ph)

73 Trng Trung hc ph thng chuyn Hong L Kha Ty Ninh Ty Ninh

(th x)

74 Trng Trung hc ph thng chuyn Nguyn Th Minh Khai

Sc Trng Sc Trng (thnh ph)

75 Trng Trung hc ph thng chuyn Nguyn Quang Diu ng Thp Cao Lnh

(thnh ph)

76 Trng Trung hc ph thng chuyn Nguyn nh Chiu ng Thp Sa c (th x)

77 Trng Trung hc ph thng chuyn Long An Long An Tn An




S 1 S GIO DC V O TO K THI TUYN SINH VO LP 10

TRNG HSP H NI TRNG THPT CHUYN I HC S PHM H NI NM HC 2013 - 2014

CHNH THC

VNG 1

Mn: Ton

Thi gian lm bi: 120 pht. Khng k thi gian giao

Cu 1: (2,5 im) 1. Cho biu thc:

3

2

a b2a a b b

ab aa bQ

3a 3b ab a a b a

vi a > 0, b > 0, a b. Chng minh gi tr ca biu thc Q khng ph thuc vo a v b. 2. Cc s thc a, b, c tha mn a + b + c = 0.

Chng minh ng thc: 2

2 2 2 4 4 4a b c 2 a b c .

Cu 2: (2,0 im)

Cho parabol (P): y = x2 v ng thng (d):

2

1y mx

2m (tham s m 0)

1. Chng minh rng vi mi m 0, ng thng (d) ct parabol (P) ti hai im phn bit.

2. Gi 1 1 2 2A x ; y , B x ; y l cc giao im ca (d) v (P).

Tm gi tr nh nht ca biu thc: 2 21 2M y y .

Cu 3: (1,5 im) Gi s a, b, c l cc s thc, a b sao cho hai phng trnh: x2 + ax + 1 = 0, x2 + bx + 1 = 0 c nghim chung v hai phng trnh x2 + x + a = 0, x2 + cx + b = 0 c nghim chung. Tnh: a + b + c.

Cu 4: (3,0 im) Cho tam gic ABC khng cn, c ba gc nhn, ni tip ng trn (O). Cc ng cao AA1, BB1, C C1 ca tam gic ABC ct nhau ti H, cc ng thng A1C1 v AC ct nhau ti im D. Gi X l giao im th hai ca ng thng BD vi ng trn (O). 1. Chng minh: DX.DB = DC1.DA1.

2. Gi M l trung im ca cnh AC. Chng minh: DH BM. Cu 5: (1,0 im) Cc s thc x, y, x tha mn:

x 2011 y 2012 z 2013 y 2011 z 2012 x 2013

y 2011 z 2012 x 2013 z 2011 x 2012 y 2013

Chng minh: x = y = z.

............. Ht ............. H v tn th sinh: ............................................................ S bo danh: ...........................

Ghi ch: Cn b coi thi khng gii thch g thm!




S GIO DC V O TO K THI TUYN SINH VO LP 10 TRNG HSP H NI TRNG THPT CHUYN I HC S PHM H NI

NM HC 2013 - 2014 CHNH THC

VNG 2

Mn: Ton


Cu 1: (2,5 im) 1. Cc s thc a, b, c tha mn ng thi hai ng thc: i) (a + b)(b + c)(c + a) = abc

ii) (a3 + b

3)(b

3 + c

3)(c

3 + a

3) = a

3b

3c

3

Chng minh: abc = 0. 2. Cc s thc dng a, b tha mn ab > 2013a + 2014b. Chng minh ng thc:

2

a b 2013 2014

Cu 2: (2,0 im) Tm tt c cc cp s hu t (x; y) tha mn h phng trnh:

3 3

2 2

x 2y x 4y

6x 19xy 15y 1

Cu 3: (1,0 im) Vi mi s nguyn dng n, k hiu Sn l tng ca n s nguyn t u tin.

S1 = 2, S2 = 2 + 3, S3 = 2 + 3 + 5, ...)

Chng minh rng trong dy s S1, S2, S3, ... khng tn ti hai s hng lin tip u l s chnh phng. Cu 4: (2,5 im) Cho tam gic ABC khng cn, ni tip ng trn (O), BD l ng phn gic ca gc ABC. ng thng BD ct ng trn (O) ti im th hai l E. ng trn (O1) ng knh DE ct ng trn (O) ti im th hai l F. 1. Chng minh rng ng thng i xng vi ng thng BF qua ng thng BD i qua trung im ca cnh AC.

2. Bit tam gic ABC vung ti B, 0BAC 60 v bn knh ca ng trn (O) bng R. Hy tnh bn knh ca ng trn (O1) theo R. Cu 5: (1,0 im) di ba cnh ca tam gic ABC l ba s nguyn t. Chng minh minh rng din tch ca tam gic ABC khng th l s nguyn. Cu 6: (1,0 im) Gi s a1, a2, ..., a11 l cc s nguyn dng ln hn hay bng 2, i mt khc nhau v tha mn:

a1 + a2 + ... + a11 = 407

Tn ti hay khng s nguyn dng n sao cho tng cc s d ca cc php chia n cho 22 s a1, a2 , ..., a11, 4a1, 4a2, ..., 4a11 bng 2012.






P N MN TON (vng 2) THI VO LP 10 TRNG THPT CHUYN HSP H NI

NM HC 2013 - 2014

Cu 1:

1.

T ii) suy ra: (a + b)(b + c)(c + a)(a2 - ab + b2)(b2 - bc + c2)(c2 - ca + a2) = a3b3c3. Kt hp vi i) suy ra: abc(a2 - ab + b2)(b2 - bc + c2)(c2 - ca + a2) = a3b3c3.

2 2 2 2 2 2 3 3 3abc 0

a ab b b bc c c ca a a b c 1

Nu abc 0 th t cc bt ng thc

2 2

2 2

2 2

a ab b ab

b bc c bc

c ca a ca

Suy ra: (a2 - ab + b

2)(b

2 - bc + c

2)(c

2 - ca + a

2) a2b2c2, kt hp vi (1) suy ra: a = b = c.

Do : 8a3 = 0 a = 0 abc = 0 (mu thun). Vy abc = 0. 2.

T gi thit suy ra:

2013 20141

b a

2

2013 2014a b a b a b

b a

2013a 2014 2013a 2014b2013 2014 2013 2 . 2014 2013 2014

b a b a

Cu 2:

Nu x = 0 thay vo h ta c:

3

2

2y 4y

15y 1

h ny v nghim.

Nu x 0, t y = tx, h tr thnh

2 33 3 3

2 2 2 2 2 2

x 1 2t 1 4tx 2t x x 4tx

6x 19tx 15t x 1 x 15t 19t 6 1

Suy ra: 3 21 2t 0;15t 19t 6 0 v 3 23 2

1 4t 162t 61t 5t 5 0

1 2t 15t 19t 6

22t 1 31t 15t 5 0

2t 1 0

1t Do t Q .

2

Suy ra: 2x 4 x 2 y 1

p s: (2; 1), (-2, -1). Cu 3:

K hiu pn l s nguyn t th n.

Gi s tn ti m m Sm-1 = k2; Sm = l

2; k, l N*.

V S2 = 5, S3 = 10, S4 = 17 m > 4. Ta c: pm = Sm - Sm-1 = (l - k)(l + k).

V pm l s nguyn t v k + l > 1 nn m

l k 1

l k p




Suy ra:

2

mm m m

p 1p 2l 1 2 S 1 S

2

(1)

Do m > 4 nn

m m2 2 2 2

2 2 2 2 2 2 m m m m

S 1 3 5 7 ... p 2 1 9

p 1 p 1 p 1 p 1 1 0 2 1 3 2 ... 8 8

2 2 2 2

(mu thun vi (1)). Cu 4:

1.

Gi M l trung im ca cnh AC.

Do E l im chnh gia ca cung AC nn EM AC. Suy ra: EM i qua tm ca ng trn (O). Di G l giao im ca DF vi (O).

Do 0DFE 90 . Suy ra: GE l ng knh ca (O). Suy ra: G, M, E thng hng.

Suy ra: 0GBE 90 , m 0GMD 90 . Suy ra t gic BDMG l t gic ni tip ng trn ng knh GD. MBD FBE .

Suy ra: BF v BM i xng vi nhau qua BD. 2.

T gi thit suy ra M l tm ng trn ngoi tip tam

gic ABC v AB =R, BC = R 3 .

Theo tnh cht ng phn gic: DA R 1

DC 3DADC R 3 3

.

Kt hp vi DA = DC = 2R.

Suy ra: 2 2DA 3 1 R DM R DA 2 3 R DE ME MD 2 2 3R

Vy bn knh ng trn (O1) bng 2 3R .

Cu 5:

Gi s a; b; c l cc s nguyn t v l di cc cnh ca tam gic ABC. t: P = a + b + c, k hiu S l din tch ca tam gic ABC. Ta c: 16S

2 = P(P - 2a)(P - 2b)(P - 2c) (1)

Gi s S l s t nhin. T (1) suy ra: P = a + b + c chn.

Trng hp 1: Nu a; b; c cng chn th a = b = c, suy ra: S = 3 (loi)

Trng hp 2: Nu a; b; c c mt s chn v hai s l, gi s a chn th a = 2.

Nu b c |b - c| 2 = a, v l.

Nu b = c th S2 = b2 - 1 (b - S)(b + S) = 1 (2) ng thc (2) khng xy ra v b; S l cc s t nhin. Vy din tch ca tam gic ABC khng th l s nguyn. Cu 6:

Ta chng minh khng tn ti n tha mn bi. Gi s ngc li, tn ti n, ta lun c: Tng cc s d trong php chia n cho a1, a2, ..., a11 khng th vt qu 407 - 11 = 396. Tng cc s d trong php chia n cho cc s 4a1, 4a2, ..., 4a11 khng vt qu 4.407 - 11 = 1617. Suy ra: Tng cc s d trong php chia n cho cc s a1, a2, ..., a11, 4a1, 4a2, ..., 4a11 khng th vt qu 396 + 1617 = 2013.

M

G

F E

D

O

C

B

A




Kt hp vi gi thit tng cc s d bng 2012. Suy ra khi chia n cho 22 s trn th c 21 php chia c s d ln nht v mt php chia c s d nh hn s chia 2 n v. Suy ra: Tn ti k sao cho ak, 4ak tha mn iu kin trn. Khi mt trong hai s n + 1; n + 2 chia ht cho ak, s cn li chia ht cho 4ak. Suy ra: (n + 1; n + 2) ak 2, iu ny khng ng. Vy khng tn ti n tha mn ra.

----- HT -----





H NI TRNG THPT CHUYN KHTN - HQG H NI NM HC 2013 - 2014

CHNH THC

Mn: Ton (vng 1)

Ngy thi: 08/06/2013


Cu 1:

1. Gii phng trnh: 3x 1 2 x 3 .

2. Gii h phng trnh:

1 1 9x y

x y 2

1 3 1 1x xy

4 2 y xy

Cu 2:

1. Gi s a, b, c l cc s thc khc 0 tha mn ng thc (a + b)(b + c)(c + a) = 8abc. Chng minh rng:

a b c 3 ab bc ca

a b b c c a 4 a b b c b c c a c a a b

2. Hi c bao nhiu s nguyn dng c 5 ch s abcde sao cho abc 10d e chia ht cho 101?

Cu 3: Cho ABC nhn ni tip ng trn (O) vi AB < AC. ng phn gic ca BAC ct (O) ti D A. Gi M l trung im ca AD v E l im i xng vi D qua O. Gi d (ABM) ct AC ti F. Chng minh rng:

1) BDM BCF. 2) EF AC.

Cu 4: Gi s a, b, c, d l cc s thc dng tha mn: abc + bcd + cad + bad = 1. Tm gi tr nh nht ca: P = 4(a3 + b3 + c3) + 9d3.






P N MN TON (vng 1) THI VO LP 10 TRNG THPT CHUYN KHTN - HQG H NI

NM HC 2013 - 2014 Cu 1:

1. Hng dn: t iu kin, bnh phng hai ln c phng trnh bc 2, nhn 2 nghim l 1, 7

4.

2. t: 1 1 1 1 1

t x ; v y tu x y xy 2y x y x xy

, ta c h phng trnh:

2

2

9t u 2u 9 2t 2u 9 2t2t 2u 92

2t 9 2t 6t 9 01 3 4tu 6t 9 0 4t 126t 9 0tu 2

4 2

u 32u 9 2t 2u 9 2t3

2t 3 t2t 3 02

2

1 3x 3 3

y 2xy 2xxy y 1 0 y 3x 0y 22 2

x 1 2x 1 02x 3x 1 01xy 3x 1 0 xy 3x 1 0y 3

x

x 1

y 2

hoc

1x

2

y 1

.

Th li, ta thy phng trnh nhn hai nghim (x; y) l 1

1; 2 ; ;12

.

Cu 2:

1. Khai trin v rt gn (a + b)(b + c)(c + a) = 8abc. Ta c: a2b + b2a + b2c + c2b + c2a + a2c = 6abc.

2 2 2 2 2 2

a ab b bc c ca 31

a b a b b c b c b c c a c a c a a b 4

ab ac ab bc ba bc ca cb ca 3

a b b c b c c a c a a b 4

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

a b b c c a 4

6abc 3

8abc 4

Lun lun ng. Suy ra: iu phi chng minh. 2. Ta c:

abc 10d e 101 101.abc abc 10d d 101 100.abc 10d e 101 abcde 101.

Vy s cc s phi tm chnh l s cc s t nhin c 5 ch s chia ht cho 101.

10000 + 100 = 101 x 100 10100 l s cc s t nhin c 5 ch s nh nht chia ht cho 101. 99999 9 = 101 x 990 99990 l s cc s t nhin c 5 ch s ln nht chia ht cho 101.

Vy s cc s t nhin c 5 ch s chia ht cho 101 l 99990 10100

1 891101

s.

Cu 3:




1. T gic AFMB ni tip AFB AMB .

M 0 0AFB BEC 180 , AMB BMD 180 BMD BED m ABDC ni tip 1 1D C

BDM BCF (g.g). Suy ra: iu phi chng minh.

2. Do 1 2A A (gt) Suy ra: D l im chnh gia cung BC.

DO BC ti trung im H ca BC.

BMD BFC 1

DABD DM BD BD DA2 .BC CF 2BH CF BH CF

M 1 2D C (chng minh trn)

BDA HCF (c.g.c) 1 1F A

M 1 2A A (gt) v

2 1A E (cng chn mtc ung DC).

1 1F E EFHC ni tip.

Cu 4: Trc ht ta chng minh vi mi x, y, y 0, ta c: x3 + y3 + z3 3xyz. (*) T chng minh 3 s hoc phn tch thnh nhn t, cc trng THPT chuyn ti TP HCM khn cho HS dng Csi. Vai tr ca a, b, c nh nhau nn gi s a = b = c = kd th P t GTNN. Khi , p dng (*), ta c:

3 3 3

2 2

3 33

3 3 2

3 33

3 3 2

3 33

3 3 2

3 3 3 3

3 2 2

1 3abca b c

k k

a b 3dabd

k k k

b c 3bdcd

k k k

c a 3dcad

k k k

2 1 33d a b c abc bcd cda dab

k k k

3 3 3 33 2 22 1 9

9d 3 a b ck k k

.

Vy ta tm k tha mn 33 2

2 13 4 4k 3k 6 0

k k

.

t

21 1

k a2 a

, ta c:

3

6 3 31 1 3 1k a a 6 x 12x 1 0 x 6 352 a 2 a

.

Lu : 3 316 35 6 35 1 k 6 35 6 352

.

Vi k xc nh nh trn, ta c: GTNN ca P bng:

22

3 3

9 36

k6 35 6 35

.

---- HT ----

1

1

1

21 F

H

E

M

D

A

CB

O





H NI TRNG THPT CHUYN KHTN - HQG H NI NM HC 2013 - 2014

CHNH THC

Mn: Ton (vng 2)

Ngy thi: 09/06/2013


Cu 1: (2,0 im) 1) Gii h phng trnh:

3 3x y 1 x y xy

7xy y x 7

2) Gii phng trnh: 2x 3 1 x 3 x 1 1 x

Cu 2: (1,5 im) 1) Tm cp s nguyn (x, y) tha mn

5x2 + 8y

2 = 20412.

2) Vi x, y l cc s thc dng tha mn x + y 1.

Tm gi tr nh nht ca biu thc: 2 21 1

P 1 x yx y

.

Cu 4: (3,5 im) Cho tam gic nhn ABC ni tip ng trn (O) c trc tm H. Gi P l im nm trn ng trn ngoi tip tam gic HBC (P khc B, C v H) v nm trong tam gic ABC. PB ct (O) ti M khc B, PC ct (O) ti N khc C. BM ct AC ti E, CN ct AB ti F. ng trn ngoi tip tam gic AME v ng trn ngoi tip tam gic ANF ct nhau ti Q khc A. 1) Chng minh rng ba im M, N, Q thng hng. 2) Gi s AP l phn gic gc MAN. Chng minh rng khi PQ i qua trung im ca BC.

Cu 5: (1,0 im) Gi s dy s thc c th t x1 x2 ... x192 tha mn cc iu kin

x1 + x2 + ... + x192 = 0 v |x1| + |x2| + ... + |x192| = 2013

Chng minh rng: 192 12013

x x96

.






P N MN TON (vng 2) THI VO LP 10 TRNG CHUYN KHTN - HQG H NI

NM HC 2013 - 2014 Cu 1:

1. Cng hai phng trnh (1) v (2) theo v, ta c: x3 + y3 + txy + y - x = 1 + y - x + xy + 7

x3 + y3 + 6xy - 8 = 0 (x + y)3 - 3xy(x + y) + 6xy - 23 = 0

(x + y - 2)[(x + y)2 + 2(x + y) + 4] - 3xy(x + y - 2) = 0

(x + y - 2)[x2 - xy + y2 + 2(x + y) + 4] = 0

x + y - 2 = 0 hoc x2 - xy + y2 + 2(x + y) + 4 = 0

Nu x + y - 2= 0 y = 2 - x thay vo (2) 7x(2 - x) + 2 - x - x - 7 = 0

7x2 - 12x + 5 = 0 (x - 1)(7x - 5) = 0

x 1 y 1

5 9x y

7 7

Th li, h phng trnh nhn nghim (x; y) l (1; 1), 5 9

;7 7

.

Nu x2 - xy + y2 + 2(x + y) + 4 = 0

4x2 - 4xy + 4y2 + 8(x + y) + 16 = 0

(x + y)2 + 8(x + y) + 16 + 3(x - y)2 = 0

(x + y + 2)2 + 3(x - y)2 = 0

(x + y + 2)2 = 3(x - y)2

x = y = -1. Thay vo (1) khng tha.

2. Gii phng trnh: 2x 3 1 x 3 x 1 1 x (1). iu kin: -1 x 1. Phng trnh (1) c vit li l:

2

2

2

x 1 x 1 1 x 1 x 2 x 1 2 0

x 1 x 1 1 1 x x 1 1 2 x 1 1 0

x 1 1 x 1 x 1 2 0

x 1 1 0

x 1 1 x 2 0

x 1 1

x 1 2 x 1. 1 x 1 x 4

x 0

1 x 1

x 0

1 x 1

x 0

Vy phng trnh c nghim duy nht l x = 0. Cu 2:

1. Trc ht ta chng minh mi s chnh phng khi chia cho 3 ch c th d 0 hoc 1. Suy ra: Tng hai s chnh phng chia ht cho 3 khi v ch khi c hai s cng chia ht cho 3.

(1) 6x2 + 9y2 - 20412 = x2 + y2 3(2x2 + 3y2 - 6804) = x2 + y2 (2)




2 2

1 12 2

2 21 1

x 3x x 9xx 3x y 3

y 3 y 3y y 9y

Thay vo (2), ta c: 2 2 2 2 2 2 2 21 1 1 1 1 1 1 13 2.9x 3.9y 6804 9x 9y 3 2x 3y 756 x y (3) 2 2

1 1 2 1 22 2

1 1 2 21 1 2 1 2

x 3 x 3x x 9xx y 3

y 3 y 3y y 9y

Thay vo (3), ta c: 2 2 2 2 2 2 2 22 2 2 2 2 2 2 23 2.9x 3.9y 756 9x 9y 3 2x 3y 84 x y (4) 2 2

2 3 2 322 2

1 1 2 22 2 3 2 3

x 3x x 9xx 3x y 3

y 3 y 3y y 9y

Thay vo (4), ta c:

2 2 2 2 2 2 2 2 2 23 3 3 3 3 3 3 3 3 33 2.9x 3.9y 84 6x 9y 28 6x 9y 28 x y 5x 8y 28 (5)

32

32 2

3 3 32

33

y 0y 0

8y 28 y 3,5 y 1y 1

y 1

Vi y3 = 0 thay vo (5) 2

35x 28 (v l, v x3 nguyn)

Vi y3 = 1 thay vo (5) 32 2

3 3

3

x 25x 8 28 x 4

x 2

Vi y3 = -1 thay vo (5) 32 2

3 3

3

x 25x 8 28 x 4

x 2

Suy ra: (x3; y3) {(2; 1), (2; -1), (-2; 1); (-2; -1)}.

V 1 2 3

1 2 3

x 3x 9x 27x

y 3y 9y 27y

nn (x; y) {(54; 27), (54; -27), (-54; 27); (-54; -27)}.

Th li phng trnh cho nhn cc nghim (x; y) {(54; 27), (54; -27), (-54; 27); (-54; -27)}. 2. p dng bt ng thc Cauchy, ta c:

11 x y 2 xy 1 4xy 4

xy

V ta cng c: 2 2 2 21 1 1 1

P 1 x y 2 1 x y 2 xyx y xy xy

M 1 15 1 1 15 1 15 2 17

xy . xy .4 2 .xyxy 16 xy 16xy 16 16xy 16 4 4

17P 2. 17

2 . Khi x = y =

1

2 th P 17 .

Vy GTNN ca P l 17 .

Cu 3:

1. Chng minh M, N, Q thng hng. Cc t gic AMEQ, ANFQ, AMCB, ANBC ni tip nn ta c: QEA QMA NMA NCA EQ / /FC .

Tng t: FQ // EB T gic EPFQ l hnh bnh hnh. Suy ra: EQF EOF BPC . Ta li c: MQE MAE MAC MBC PBC




NQF NAF NAB NCB PCB 0EQM EQF FQN PBC BPC PCB 180 .

Suy ra: M, Q, N thng hng. 2. Chng minh PQ qua trung im ca BC. Ke ng cao CI, BJ ca tam gic ABC. EF ct PQ ti G. Do t gic AMEQ, ANFQ ni tip v QEPH l hnh bnh hnh nn ta c:

QAM QEP QFP QAN . Do AP l

phn gic ca MAN . Suy ra: A, Q, P thng hng. Gi giao ca AP vi BC l K. Ta c:

IHJ BHC BPC FPE IHJ FPE

M 0IHJ IAJ 180 0 0FPE IAJ 180 FPE FAE 180

Suy ra: FPEA ni tip. EFP EAP EAQ EMQ EMN BMN BCN EF / /BC FG AG GE

BK AK KC

M FG = GE BK = KC PQ l trung im ca K ca BC. Cu 4:

Ta chng minh bi ton: 1 2 na a ... a tha mn 1 2 3 n

1 2 3 n

a a a ... a 0

a a a ... a 1

th n 1

2a a

n .

T iu kin trn, ta suy ra: C k N sao cho 1 2 k k 1 na a ... a 0 a ... a

1 2 k1 2 k k 1 n

1 2 k k 1 nk 1 n

1a a ... aa a ... a a ... a 0 2

1a a ... a a ... a 1a ... a

2

M

1 2 k 1 k 1 n n

n 1 2

1 1a a ... a a ; a ... a a

2k 2k

1 1 n n 2a a

2k 2 n k 2k n k nk n k2

2

Bi ton ph c chng minh.

T (I) suy ra:

1921 2

1921 2

xx x... 0

2013 2013 2013

xx x... 0

2013 2013 2013

p dng bi ton trn, ta c:

192 1192 1

x x 2 2013x x

2013 2013 192 96 (iu phi chng minh)

---- HT ----





H NI TRNG THPT CHUYN NGOI NG - HNN - HQG H NI NM HC 2013 - 2014

CHNH THC

thi mn ton ca trng THPT chuyn ngoi ng - HNN - HQG H Ni l thi ca trng chuyn KHTN - HQG H Ni.







H NI TRNG THPT CHUYN H NI - AMSTERDAM NM HC 2013 - 2014

CHNH THC

Mn: Ton


Cu 1:

1) Tm cc s t nhin n 72013 + 3n c ch s hng n v l 8.

2) Cho a, b l cc s t nhin ln hn 2 v p l s t nhin tha mn 2 2

1 1 1= +

p a b.

Chng minh p l hp s.

Cu 2:

1) Tm cc s nguyn x, y tha mn: x

2 3y2 + 2xy 2x + 6y 8 = 0. 2) Gii h phng trnh:

2 2

2 2

2x xy 3y 2y 4 0

3x 5y 4x 12 0

Cu 3: Cho a, b l cc s thc tha mn: a + b + 4ab = 4a2 + 4b2. Tm gi tr ln nht ca biu thc: A = 20(a3 + b3) 6(a2 + b2) + 2013.

Cu 4: Cho tam gic ABC khng phi l tam gic cn. ng trn (O) tip xc vi BC, AC, AB ln lt ti M, N, P. ng thng NP ct BO, CO ln lt ti E v F.

1) Chng minh rng OEN v OCA bng nhau hoc b nhau.

2) Bn im B, C, E, F thuc mt ng trn.

3) Gi K l tm ng trn ngoi tip OEF. Chng minh ba im O, M, K thng hng.

Cu 5: Trong mt phng cho 6 im A1, A2, ..., A6, trong khng c ba im no thng hng v trong ba im lun c hai im c khong cch nh hn 671. Chng minh rng trong su im cho lun tn ti ba im l ba nh ca mt tam gic c chu vi nh hn 2013.


Ghi ch: Cn b coi thi khng gii thch gi thm!





H NI TRNG THPT CHU VN AN H NI NM HC 2013 - 2014

CHNH THC

Mn: Ton

Thi gian lm bi: 150 pht. ( thi tuyn sinh vo lp 10 THPT ca TP H Ni)

Cu I: (2,0 im) Vi x > 0, cho hai biu thc: 2 x

Ax

v

x 1 2 x 1B

x x x

.

1) Tnh gi tr ca biu thc A khi x = 64. 2) Rt gn biu thc B.

3) Tnh x A 3

B 2

Cu II: (2,0 im) Gii bi ton bng cch lp phng trnh: Qung ng t A n B di 90 km. Mt ngi i xe my t A n B. khi n B, ngi ngh 30 pht ri quay tr v A vi vn tc ln hn vn tc lc i l 9 km/h. Thi gian k t lc bt u i t A n lc tr v n A l 5 gi. Tnh vn tc xe my lc i t A n B.

Cu III: (2,0 im)

1) Gii h phng trnh:

3 x 1 2 x 2y 4

4 x 1 x 2y 9

2) Cho parabol (P): 21

y x2

v ng thng (d): 21

y mx m m 12

.

a) Vi m = 1, xc nh ta giao im A, B ca (d) v (P). b) Tm cc gi tr ca m (d) ct (P) ti hai im phn bit c honh x1, x2 sao cho:

1 2x x 2 .

Cu IV: (3,5 im) Cho ng trn (O) v im A nm bn ngoi (O). K hai tip tuyn AM, AN vi ng trn (O). Mt ng thng d i qua A ct ng trn (O) ti hai im B v C (AB < AC, d khng i qua tm O). 1) Chng minh t gic AMON ni tip. 2) Chng minh: AN2 = AB.AC. Tnh di on thng BC khi AB = 4cm, AN = 6cm. 3) Gi I l trung im BC. ng thng NI ct ng trn (O) ti im th hai T. Chng minh MT//AC.

4) Hai tip tuyn ca ng trn (O) ti B v C ct nhau ti K. Chng minh K thuc mt ng thng c nh khi d thay i v tha mn iu kin u bi.

Cu V: (0,5 im) Vi a, b, c l cc s dng tha mn iu kin: a + b + c + ab + bc + ca = 6abc. Chng minh:

2 2 2

1 1 13

a b c


Ghi ch: Cn b coi thi khng gii thch g thm! (im chun ca trng nm 2013 l 52,0 im.)




P N MN TON THI VO LP 10 TRNG THPT CHU VN AN - H NI

(K THI TUYN SINH VO LP 10 NM 2013 - 2014) Cu 1:

1) Vi x = 64, ta c: 2 64 2 8 5

A8 464

2)

x 1 x x 2 x 1 x x x 2x 1 x 2

B 1x x x x 1 x 1x x x

3) Vi x > 0, ta c:

A 3 2 x 2 x 3 x 1 3

: 2 x 2 3 x x 2 0 x 4. Do x 0B 2 2 2x x 1 x

Cu 2:

t: x (km/h) l vn tc i t A n B. Vy vn tc i t B n A l x + 9 (km/h) Do gi thit, ta c:

290 90 1 10 10 1

5 x x 9 20 2x 9 x 31x 180 0 x 36x x 9 2 x x 9 2

(nhn)

Cu 3:

1) H phng trnh tng ng vi:

3x 3 2x 4y 4 5x 4y 1 5x 4y 1 11x 11 x 1

4x 4 x 2y 9 3x 2y 5 6x 4y 10 6x 4y 10 y 1

2) Vi m = 1, ta c phng trnh honh giao im ca (P) v (d) l

2 21 3

x x x 2x 3 0 x 1hay x 3 Doa b c 02 2

Ta c:

x = - 1 1

y2

v x = 3 9

y2

.

Vy ta giao im ca A v B l 1

1; 2

v

93;

2

.

b) Phng trnh honh giao im ca (P) v (d) l:

2 2 2 21 1

x mx m m 1 x 2mx m 2m 2 0 *2 2

(d) ct (P) ti hai im phn bit x1, x2 th phng trnh (*) phi c 2 nghim phn bit.

Khi : ' = m2 - m2 + 2m + 2 > 0 m > -1.

Khi m > -1, ta c: 22 2

1 2 1 2 1 2 1 2 1 2x x 2 x x 2x x 4 x x 4x x 4

2 21

4m 4 m 2m 2 4 8m 4 m2

Cu 4:

1) Xt t gic AMON c hai gc i

0

0

ANO 90

AMO 90

Nn l t gic ni tip.

2) V ABM ACM nn ta c: AB.AC = AM2 = AN2 = 62 = 36.

2 26 6

AC 9 cmAB 4

BC = AC - AB = 9 - 4 = 5(cm)




3)

1MTN MON AON2

(cng chn cung MN trong

ng trn (O)) v AIN AON . (Do 3 im M, I, N cng nm trn ng trn ng knh AO v cng chn cung 900)

Vy AIN MTI TIC nn MT//AC (do c hai gc so le bng nhau).

4) Xt AKO c AI KO. H OQ vung gc vi AK. Gi H l giao im ca OQ v AI th H l trc tm

ca AKO nn KH AO .

V MN AO nn ng thng KMHNAO nn

KM AO. Vy K nm trn ng thng c nh MN khi BC di chuyn.

Cu 5:

T gi thit cho, ta c:

1 1 1 1 1 16

ab bc ca a b c .

Theo bt ng thc Cauchy, ta c:

2 2 2 2 2 2

1 1 1 1 1 1 1 1 1 1 1 1; ;

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

2 2 2

1 1 1 1 1 1 1 1 11 ; 1 ; 1

2 a a 2 b b 2 c c

Cng cc bt ng thc trn, v theo v, ta c:

2 2 2 2 2 2 2 2 2

3 1 1 1 3 3 1 1 1 3 9 1 1 16 6 3

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

(pcm)

----- HT -----

Q

H

P

K

T

IC

O

B

N

M

A





H NI TRNG THPT SN TY H NI NM HC 2013 - 2014

CHNH THC

S dng thi TUYN SINH VO LP 10 nm hc 2013 - 2014 ca TP. H Ni xt tuyn. Cng l thi vo lp CHU VN AN H Ni

(im chun ca trng nm 2013 l 46,0 im.)





H NI TRNG THPT CHUYN NGUYN HU NM HC 2013 - 2014

CHNH THC

Mn: Ton


( THI NY CNG L THI VO LP 10 CHUYN TON H NI - AMSTERDAM

NM 2013 - 2014) Cu 1:

1. Tm cc s t nhin n 72013 + 3n c ch s hng n v l 8.

2. Cho a, b l cc s t nhin ln hn 2 v p l s t nhin tha mn 2 2

1 1 1= +

p a b.

Chng minh p l hp s.

Cu 2:

1. Tm cc s nguyn x, y tha mn: x

2 3y2 + 2xy 2x + 6y 8 = 0. 2. Gii h phng trnh:

2 2

2 2

2x xy 3y 2y 4 0

3x 5y 4x 12 0

Cu 3: Cho a, b l cc s thc tha mn: a + b + 4ab = 4a2 + 4b2. Tm gi tr ln nht ca biu thc: A = 20(a3 + b3) 6(a2 + b2) + 2013.

Cu 4: Cho tam gic ABC khng phi l tam gic cn. ng trn (O) tip xc vi BC, AC, AB ln lt ti M, N, P. ng thng NP ct BO, CO ln lt ti E v F.

1. Chng minh rng OEN v OCA bng nhau hoc b nhau.

2. Bn im B, C, E, F thuc mt ng trn.

3. Gi K l tm ng trn ngoi tip OEF. Chng minh ba im O, M, K thng hng.

Cu 5: Trong mt phng cho 6 im A1, A2, ..., A6, trong khng c ba im no thng hng v trong ba im lun c hai im c khong cch nh hn 671. Chng minh rng trong su im cho lun tn ti ba im l ba nh ca mt tam gic c chu vi nh hn 2013.







TP. H CH MINH TRNG PTNK - HQG TP H CH MINH NM HC 2013 - 2014

CHNH THC

Mn: Ton


Cu 1:

Cho phng trnh: x2 - 4mx + m2 - 2m + 1 = 0 (1) vi m l tham s. a) Tm m phng trnh c hai nghim x1, x2 phn bit. Chng minh rng khi hai nghim khng th tri du nhau.

b) Tm m sao cho: 1 2x x 1

Cu 2:

Gii h phng trnh:

2

2

2

3x 2y 1 2z x 2

3y 2z 1 2x y 2

3z 2x 1 2y z 2

Cu 3: Cho x, y l hai s khng m tha mn: x3 + y3 x - y. a) Chng minh rng: y x 1. b) Chng minh rng: x3 + y3 x2 + y2 1.

Cu 4: Cho M = a

2 + 3a + 1, vi a l s nguyn dng.

a) Chng minh rng mi c s ca M u l s l. b) Tm a sao cho M chia ht cho 5. Vi nhng gi tr no ca a th M l ly tha ca 5.

Cu 5:

Cho ABC c 0A 60 . ng trn (I) ni tip tam gic tip xc vi cc cnh BC, CA, AB ln lt ti D, E, F. ng thng ID ct EF ti K, ng thng qua K song song vi BC ct AB, AC ln lt ti M, N. a) Chng minh rng: IFMK v IMAN l t gic ni tip. b) Gi J l trung im BC. Chng minh A, K, J thng hng. c) Gi r l bn knh ng trn (I) v S l din tch t gic IEAF. Tnh S theo r v chng minh:

IMNS

S4

Cu 6:

Trong mt k thi, 60 hc sinh phi gii 3 bi ton. Khi kt thc k thi, ngi ta nhn thy rng: Vi hai th sinh bt k lun c t nht mt bi ton m c hai th sinh gii c. Chng minh rng: a) Nu c mt bi ton m mi th sinh khng gii c th phi c mt bi ton khc m mi th sinh u gii c. b) C mt bi ton m c t nht 40 th sinh u gii c.


Gh ch: Cn b coi thi khn gii thch g thm!




P N MN TON THI VO LP 10 TRNG PTNK - HQG TP H CH MINH

NM HC 2013 - 2014 Cu 1:

a) Phng trnh c hai nghim phn bit 1 2

x ,x 2 2' 4m m 2m 1 0

23m m 3m 1 0 m 3m 1 3m 1 0

3m 1 m 1 0 1

1m va m > -13m 1 0 va m 1 0 m3

33m 1 0 va m 1 0 1

m 1m < va m < -1

3

Khi : 2

2

1 2x .x m 2m 1 m 1 0

Do 1 2

x ;x khng th tri du.

b) Phng trnh c hai nghim khng m 1 2

x ;x

1 2

2

1 2

1m hoac m 1 (ap dung cau a)

' 0 3

S x x 0 4m 0

P x .x 0 m 1 0

1

m

3

Ta c: 2

1 2 1 2 1 2x x 1 x x 2 x x 1 4m 2 m 1 1

4m 14m 2 m 1 1 m 1

2

4m 10

2

4m 1m 1

2

1 4mm 1

2

1m

44m 1

11m (thch hp)2m 2 4m 1 m

222m 2 1 4m

1m

2

Vy 1

m

2

l gi tr cn tm.

Cu 2:

Ta c: 2 2 23x 2y 1 3y 2z 1 3z 2x 1 2z x 2 2x y 2 2y z 2 2 2 2

3x 2y 1 3y 2z 1 3z 2x 1 2zx 4z 2xy 4x 2yz 4y

2 2 2 2 2 2 2 2 2

2 2 2 2 2 2

x 2xy y x 2zx z y 2yz z x 2x 1 y 2y 1 z 2z 1 0

x y x z y z x 1 y 1 z 1 0

2 2 2 2 2 2

x y x z y z x 1 y 1 z 1 0

x y;x z;y z;x 1;y 1;z 1 x y z 1

Th li, ta c: x;y;z 1;1;1 l nghim ca h phng trnh cho.




Cu 3:

a) Ta c: x 0;y 0 v 3 3x y x y .

Do : 3 3x y x y 0 . Nn x y 0 x y

Ta cng c : 3 3 3 3 2 2x y x y x y x xy y Nn 2 2x y x y x xy y Nu x = y th 3 3x y 0 . Ta c : x = y = 0. Nn y x 1

Nu x y th t 2 2x y x y x xy y ta c : 2 21 x xy y M 2 2 2x xy y x . Nn 21 x . M x 0 . Nn 1 x

Vy y x 1

b) 0 y x 1 nn 3 2 3 2y y ;x x . Do : 3 3 2 2x y x y

V 2 21 x xy y v 2 2 2 2x xy y x y . Do : 2 2x y 1

Vy 3 3 2 2x y x y 1

Cu 4:

a) 2 2M a 3a 1 a a 2a 1 a a 1 2a 1 l s l (v a, a + 1 l hai s nguyn dng lin

tip nn a a 2 1 ) Do mi c c M u l s l.

b) 2

2 2

M a 3a 1 a 2a 1 5a a 1 5a

Ta c: M 5; 5a 5 . Do : 2

a 1 5 . Nn a 1 5

Ta c : a chia cho 5 d 1, tc a 5k 1 k N

t 2 n *a 3a 1 5 n N ( *n N v do a 1 nn 2a 3a 1 5 ) Ta c : n5 5 theo trn ta c : a 5k 1 k N

Ta c : 2

n

5k 1 3 5k 1 1 5 2 n25k 10k 1 15k 3 1 5

n25k k 1 5 5 *

Nu n 2 ta c : n 25 5 , m 225k k 1 5 ; 5 khng chia ht cho 25 : v l.

Vy n = 1. Ta c : 25k k 1 0;k N . Do : k = 0. Nn a = 1. Cu 5:

a) Ta c : MN // BC (gt), ID BC ((I) tip xc vi BC ti D) 0

ID MN IK MN IKM IKN 90 0 0 0IFM IKM 90 90 180

T gic IFMK ni tip.

Mt khc : 0IKN IEN 90 T gic IKEN ni tip.

Ta c : IMF IKF (T gic IFMK ni tip) ; IKF ANI (T gic IKEN ni tip).

IMF ANI T gic IMAN ni tip.

b) Ta c :




IMK IFK T giac IFMK noi tiep

INK IEK T giac IKEN noi tiep

Mt khc : IE = IF (= r)

IEF cn ti I.

IMN cn ti I c IK l ng cao.

IK l ng trung tuyn ca IMN

K l trung im ca MN. MN 2.MK M BC = 2.BJ (J l trung

im ca BC)

Do : MN 2.MK MK

BC 2.BJ BJ

Mt khc: ABC c MN // BC

AM MN

AB BC

(H qu ca nh l Thales)

Ta c: AM MK MN

AB BJ BC

Xt AMK v ABJ , ta c:

AMK ABJ hai goc ong v va MN // BCAM MK

AB BJ

AMK ABJ c g c MAK BAJ Hai tia AK, AJ trng nhau. Vy ba im A, K, J thng hng. c) AE, AF l cc tip tuyn ca ng trn (I)

AE = AF, AI l tia phn gic ca EAF

AEF cn ti A c 0EAF 60 (gt) AEF u. EF = AE = AF.

AEF u c AI l ng phn gic.

AI l ng cao ca AEF

AI EF 1

S AI.EF

2

IAE vung ti E AE = IE.cotIAE ; IE = AI.sin. IAE

0

0

rAE r.cot 30 3.r;AI 2r

sin30

Vy EF = AE = 3.r

Vy 21 1

S .AI.EF .2r. 3.r 3.r (vdt)

2 2

Gi H l giao im ca AI v EF.

Ta c: IH EF, H l trung im ca EF v 0HIF 60 .

J

NMK

CB

A

F

E

D

I




IHF vung ti H 01

IH IF.cosHIF r.cos60 .r2

Do : 2

IEF

1 3.rS .IH.EF

2 4

(vdt)

Xt IMN v IEF , ta c: IMN IFE;INM IEF

Do : IMN IEF (g.g) 2

IMN

IEF

S IM

S IF

. M IF FM

IMIM IF 1

IF

Do : IMNIMN IEF

IEF

S

1 S S

S

Ta c: 2

2

IEF IMN IEF

3.rS 3.r ;S ;S S

4

Vy IMN

SS

4

Cu 6: Gi ba bi ton l A, B, C. a) Khng mt tnh tng qut, gi s mi th sinh u khng gii c bi ton A. Nu mi th sinh u khng gii c bi ton B th t gi thit ta c mi th sinh u gii c bi ton C.

Nu mi th sinh u gii c bi ton B v bi ton C th ta c mi th sinh u gii c bi ton B; bi ton C.

Nu c mt th sinh ch gii c mt bi ton, gi s gii c bi ton B. Xt hc sinh ny vi tt c cc hc sinh cn li. Theo gi thit, c mi th sinh u gii c bi ton B. Vy nu c mt bi ton m mi th sinh u khng gii c th phi c mt bi ton khc m mi th sinh u gii c. b) Theo gi thit ta c mi th sinh u gii c t nht mt bi ton. Nu c mt th sinh ch gii ng mt bi ton. Xt hc sinh ny vi tt c cc hc sinh cn li, ta c mi th sinh u gii c bi ton . Ta ch cn xt trng hp m mi th sinh gii c t nht hai bi ton. Gi s th sinh gii c A, B m khng gii c C l x, s th sinh gii c B, C m khng gii c A l y, s th sinh gii c A, C m khng gii c B l z, s th sinh gii c c A, B, C l t

(x, y, z, t N ) Ta c: x + y + z + t = 60 (1)

Gi s c iu tri vi kt lun ca bi ton. Ta c: x + z + t < 40; x + y + t < 40; y + z + t < 40

Do : x + z + t + x + y + t + y + z + t < 40 + 40 + 40 2 x y z t t 120 . Kt hp (1) ta c : t < 0. iu ny v l. iu gi s trn l sai. Vy c mt bi ton m c t nht 40 th sinh gii c.

----- HT -----





TP. H CH MINH TRNG THTH - HSP TP. H CH MINH NM HC 2013 - 2014

CHNH THC

Mn: Ton


thi ny c 01 trang

Cu 1: Cho phng trnh: 2 2x - (2m-3)x + m - 2m + 2 = 0 , (m l tham s)

1) Tm m phng trnh c mt nghim l -1. Tm nghim cn li.

2) Tm m phng trnh c hai nghim x1, x2 tha: 2 2

1 2 1 1x + x + x + x = 2 .

Cu 2: Cho hm s:

2

xy = -

2 (P) v y = mx - 4 (D), vi m 0.

1) Khi m = 1, hy v (P) v (D) trn cng trn mt mt phng ta Oxy. Tm ta giao im ca (D) v (P) bng php tnh.

2) Tm m (P), (D) v (D): 1

y = x +2

ng quy.

Cu 3: Cho biu thc: 3x +5 x -11 x - 2 2

P = - + -1x + x - 2 x -1 x + 2

, vi x 0 v x 1.

1) Rt gn P. 2) Tm x P nhn gi tr nguyn.

Cu 4: Gii h phng trnh:

2

4

x 4x y 0

x 2 5y 16

Cu 5: Cho tam gic ABC nhn (AB < AC) c ng cao AH. V ng trn (O) ng knh AB ct AC ti N. Gi E l im i xng ca H qua AC, EN ct AB ti M v ct (O) ti im th hai D.

1) Chng minh: AD = AE.

2) Chng minh HA l phn gic ca MHN . 3) Chng minh: a) 5 im A, E, C, M, H thuc ng trn (O1). b) 3 ng thng CM, BN, AH ng quy. 4) DH ct (O1) ti im th hai Q. Gi I, K ln lt l trung im ca DQ v BC. Chng t I thuc ng trn (AHK).







TP. H CH MINH TRNG THPT CHUYN L HNG PHONG TP. HCM NM HC 2013 - 2014

CHNH THC

Mn: Ton (chung)

Ngy thi: 22/06/2013


thi ny c 01 trang

Cu 1: (2,0 im)

1. Gii phng trnh: x 2x 2 5x 9

2. Cho x, y, z i mt khc nhau tha mn 1 1 1

0x y z .

Tnh gi tr biu thc: 2 2 2

yz zx xy

x 2yz y 2zx z 2xy

.

Cu 2: (1,5 im) Cho phng trnh: x2 - 5mx - 4m = 0. 1. nh m phng trnh c hai nghim phn bit. 2. Gi x1, x2 l hai nghim ca phng trnh.

Tm m biu thc: 22

2 1

2 2

1 2

x 5mx 12mm

x 5mx 12m m

t gi tr nh nht.

Cu 3: (1,5 im) Cho tam gic ABC c BC l cnh di nht. Trn BC ly hai im D v E sao cho BD = BA, CE = CA. ng thng qua D song song vi AB ct AC ti M. ng thng qua E song song vi AC ct AB ti N. Chng minh rng: AM = AN.

Cu 4: (1,0 im)

Cho x, y l hai s dng tha mn: x + y = 1. Chng minh rng: 2 8x

3 3x 2 7y

.

Cu 5: (1,0 im) T mt im A bn ngoi ng trn (O) v cc tip tuyn AB, AC v ct tuyn AEF (EF khng i qua O, B v C l cc tip im). Gi D l im i xng ca B qua O. DE, DF ln lt ct AO ti M v N. Chng minh rng: 1. Tam gic CEF ng dng vi tam gic CMN. 2. OM = ON.

Cu 6: (1,5 im)

Ch s hng n v trong h thp phn ca s M = a2 + ab + b2 l 0 (a, b N*). 1. Chng minh rng M chia ht cho 20. 2. Tm ch s hng chc ca M.


Ghi ch: Cn b coi thi khng gii thch g thm! imchun chuyn Ton: NV1: 38.5 im; NV2: 39.25 im.




P N MN TON THI TUYN SINH VO LP 10

TRNG THPT CHUYN L HNG PHONG - TP HCM NM HC 2013 - 2014

Cu 1:

1. iu kin: x 1.

2 2

2 2

2

2

x 2x 2 5x 9

2x 2x 2 10x 18

x 2x 2x 2 2x 2 x 8x 16

x 2x 2 x 4

x 2x 2 x 4 0

x 2x 2 x 4 0

2x 2 4 2x 1 x 2

2x 2 16 16x 4x

2x 9x 9 0

x = 3 (loi); 3

x2

(nhn)

Vy phng trnh c nghim 3

x2

.

2. iu kin: xyz 0. T gi thit, suy ra: xy + yz + zx = 0.

yz = -xy - zx. Do :

2 2yz yz yz

x 2yz x yz xy zx x y z x

Chng minh tng t, ta c:

2yz yz

1x 2yz x y z x

(quy ng)

Cu 2:

1. Phng trnh c hai nghim phn bit khi v ch khi:

= (5m)2 + 16m 0

m(25m - 16) 0

m 0

16m

25

2. Ta c: 2 2

1 1 1 1

2 2

2 2 2 2

x 5mx 4m 0 x 5mx 4m

x 5mx 4m 0 x 5mx 4m

Thay vo biu thc trn, ta c:




22

2 1

2 2

1 2

21 2

2

1 2

2 2

2 2

x 5mx 12mmA

x 5mx 12m m

5m x x 16mm

5m x x 16m m

m 25m 16m 2

25m 16m m

minA 2 t c khi 2

m3

.

Cu 3:

N

M

E DCB

A

V BC l cnh ln nht nn D, E thuc cnh BC. p dng nh l thales vo cc tam gic ABC, ta c:

AM BD

AC BC m AC = CE nn

BD.CEAM

BC

AN CE

AB BC m AB = BD nn

BD.CEAN

BC

Vy AM = AN.

Cu 4:

Ta c: x + y = 1 y = 1 - x.

2 8x

3 3x 2 7y

2

2

83 3x 2 7

1 x

83 3x 1 18 1 x 24

1 x

Theo bt ng thc C si cho hai s dng 18 1 x v 8

1 x, ta c:

2 8 8 8

3 3x 1 18 1 x 18 1 x 2 18 1 x . 2 18.8 241 x 1 x 1 x

ng thc xy ra khi 3(3x - 1)2 = 0 3x - 1 = 0 1

x3

.

Khi 1 2

x y .3 3

Cu 5:




N

M

D

F

E

C

B

A O

1. Ta c:

DEC DBC OAC. Suy ra: T gic ACNE ni tip CMD CAE . CFD CBD CAN . Suy ra: T gic ACNF ni tip CND CAE .

Suy ra: CND CMD. Do hnh thang CMND (MN//CD) ni tip c nn l hnh thang cn.

Suy ra: CNM EDC CFE (1)

Ta c: 0 0CMN 180 CMA 180 CEA CEF (2) T (1) v (2), suy ra: CEF CMN. 2.

T gic CMND l hnh thang cn nn CNM NMD .

M CNM BNM nn BNM NMD . Suy ra: BN//DM (3)

M DM = CN = BN (4)

Nn t gic BMDN l hnh bnh hnh. Suy ra hai ng cho MN v BD ct nhau ti O l trung im ca mi ng. Vy OM = ON. Cu 6:

1. V ch s tn cng l 0 nn M5. Xt cc trng hp: (1) C hai s a, b u l. Suy ra: a

2, b

2, ab u l hay M l (v l, v M tn cng l 0) (2) Mt trong hai s a, b c mt s l, mt s chn. Khng mt tnh tng qut, gi s s l l a, s chn l b. Suy ra: a

2 l, b2 v ab chn hay M l (v l, v M tn cng l 0)

Do c hai a, b chn.

Khi : 2 2a 4; b 4; ab 4 hay M 4 . Vy M 4.5=20 (v (4 ,5) = 1). 2. Ta c: 2 2 3 3 3 3 3 3 6 6a ab b a b a b 5 a b a b a b 5 Ta li c: 6 2 2 2 2a a a a 1 a 1 a 1 a a 2 a 1 a a 1 a 2 5a a 1 a 1 5 vi aN (v tch ca 5 s t nhin lin tip lun chia ht cho 5).

Tng t: 6 2b b 5 b N

6 6 6 2 6 2 2 2a b a a b b a b 5

2 2 3 3a b a b a b ab a b 5

2 2ab a b a b ab a 2ab b 5 (1)




M 2 2ab.M ab a ab b 5 (2)

T (1) v (2), suy ra: 22 2ab.3ab 3a b 5 ab 5 ab 5

Ta c: M = a2 + ab + b

2 35 b.M ab a b b 5

M ab(a + b) 5 do ab 5 b3 5 b 5 . a2 = M - b(a + b) 5 a 5 . M 5 . M theo cu 1, ta c: M 4 . Ta li c: (4, 25) = 1 nn M 4.25 100 . Vy ch s hng chc ca M l 0.

----- HT -----





TP. H CH MINH TRNG THPT NGUYN THNG HIN TP. HCM NM HC 2013 - 2014

CHNH THC

Mn: Ton

Thi gian lm bi: 150 pht. y l chnh thc ca thi tuyn sinh vo lp 10 THPT

nm hc 2013 - 2014 ca TP. H Ch Minh

Cu 1: (2,0 im) Gii cc phng trnh v h phng trnh sau: a) x

2 - 5x + 6 = 0

b) x2 - 2x - 1 = 0

c) x4 + 3x

2 - 4 = 0

2x y 3

d)x 2y 1

Cu 2: (1,5 im) a) V th (P) ca hm s y = x2 v ng thng (D): y = -x + 2 trn cng mt h trc ta . b) Tm ta cc giao im ca (P) v (D) cu trn bng php tnh. Cu 3: (1,5 im) Thu gn cc biu thc sau:

x 3 x 3

A .x 9x 1 x 3

vi x 0; x 9

2 2

B 21 2 3 3 5 6 2 3 3 5 15 15

Cu 4: (1,5 im) Cho phng trnh: 8x2 - 8x + m2 + 1 = 0 (*) (x l n s)

a) nh m phng trnh (*) c nghim 1

x2

.

b) nh m phng trnh (*) c hai nghim x1, x2 tha iu kin: 4 4 3 3

1 2 1 2x x x x

Cu 5: (3,5 im) Cho tam gic ABC khng c gc t (AB < AC), ni tip ng trn (O; R). (B, C c nh, A i ng trn cung ln BC). Cc tip tuyn ti B v C ct nhau ti M. T M k ng thng song song vi AB, ng thng ny ct (O) ti D v E (D thuc cung nh BC), ct BC ti F, ct AC ti I.

a) Chng minh rng: MBC BAC . T suy ra MBIC l t gic ni tip. b) Chng minh: FI.FM = FD.FE. c) ng thng OI ct (O) ti P v Q (P thuc cung nh AB). ng thng QF ct (O) ti T (T khc Q). Chng minh ba im P, T, M thng hng. d) Tm v tr im A trn cung ln BC sao cho tam gic IBC c din tch ln nht.


Ghi ch: Cn b coi thi khng gii thch g thm! (im chun vo trng l: NV1: 38,25 im; NV2: 39,25 im; NV3: 40,25im)




P N MN TON THI VO LP 10 THPT CA TP. H CH MINH

NM HC 2013 - 2014 Cu 1:

1a) x2 - 5x + 6 = 0

= 1. Suy ra: x1 = 1; x2 = 3. 1b) x

2 - 2 x - 1 = 0

' = 2. Suy ra: 1 1x 1 2; x 1 2

1c) x4 + 3x

2 - 4 = 0.

t: t = x2, (t 0) Phng trnh tr thnh: t2 + 3t - 4 = 0.

= 25 > 0. t1 = -4 (loi) v t2 = 1 (nhn)

Vi t = 1 x = 1.

12x y 3

d)x 2y 1

Gii h phng trnh trn, ta c: x 1

y 1

.

Cu 2:

2a) V (P) v lp bng gi tr ng. V (D) v lp bng gi tr ng. 2b) Phng trnh honh giao im gia (P) v (D) l: x

2 = - x + 2

x2 + x - 2 = 0

x = 1 hoc x = -2.

Vi x = 1 y = 1

Vi x = - 2 y = 4. Cu 3:

3a)

x 3 x 3A .

x 9x 1 x 3

x 9 x 3.

x 9x 3 x 3

1

x 3

3b)

2 2

B 21 2 3 3 5 6 2 3 3 5 15 15

2 22 2 2 2

2 2

3 1 5 1 3 1 5 121 6 15 15

2 2 2 2

3 5 3 521 6 15 15

2 2

15 4 15 15 15 60




Cu 4:

4a) Ta c: 8x2 - 8x + m

2 + 1 = 0.

Phng trnh c nghim l 21

x m 1 m 1.2

4b) Ta c: ' = 8 - 8m2. Phng trnh c nghim khi v ch khi 8 - 8m2 0.

Theo nh l Vi - t, ta c: x1 + x2 = 1 v x1x2 = 2m 1

8

.

4 4 3 3

1 2 1 2

3 3

1 1 2 2

1 2 1 2 1 2

1 2

x x x x

x 1 x x 1 x 0

x x x x x x 0

x x

m 1.

Cu 5:

5a)

MBC BAC (cng chn BC )

MIC BAC (ng v)

MBC MIC T gic MBIC ni tip. 5b)

IFC BFM FI.FM = FB.FC BFD EFC FD.FE = FB.FC Vy FI.FM = FD.FE. 5c)

0PTQ 90 (gc ni tip na ng trn) BFT QFC FT.FQ = FB.FC M FI.FM = FB.FC FI.FM = FT.FQ

MFT QFI.

Suy ra: MTQ MIQ (1) T gic MBOC v t gic MBIC ni tip nn 5 im M, B, O, I, C cng thuc ng trn ng knh OM.

Suy ra: 0MIQ MIO=90 (2)

T (1) v (2) 0MTQ 90 . 0 0 0PTQ MTQ 90 90 180 P, T, M thng hng. K BH AC.

IBC

1 1S BH.IC IB.IC.sin BIA

2 2 .

Do 0BIA 180 2BAC khng i nn SIBC ln nht khi IB.IC ln nht.

IB.IC = IA.IC

2 22IA IC AC R

2 4

Du "=" xy ra khi IA = IC v A i xng vi C qua tm O.

--- HT ---

T

F

P

Q

I

H

E

K

M

CB

A

O





TP. H CH MINH TRNG THPT GIA NH TP. HCM NM HC 2013 - 2014

CHNH THC

y l chnh thc ca thi tuyn sinh vo lp 10 THPT nm hc 2013 - 2014 ca TP. H Ch Minh

im chun lp chuyn:

LP CHUYN NGUYN VNG 1 NGUYN VNG 2

Ting Anh 34.5 35.25

Ho hc 31 31.25

Vt l 29.75 30

Ton 30.75 31

Ng vn 32.5 33.5


Ghi ch: Cn b coi thi khng gii thch g thm! (im chun vo trng: NV1: 34,50 im; NV2: 35,50 im; NV3: 36,50 im)





TP. H CH MINH TRNG THPT CHUYN TRN I NGHA NM HC 2013 - 2014

CHNH THC

thi vo lp 10 trng THPT chuyn Trn i ngha l thi vo lp 10 trng THPT chuyn L Hng Phong TP H Ch Minh nm hc 2013 - 2014.

im chun lp chuyn:

LP CHUYN NGUYN VNG 1 NGUYN VNG 2

Ting Anh 36.5 37.25

Ho hc 34.25 35

Vt l 35 35.5

Sinh hc 34.75 35.5

Ton 34.75 35.5

Ng vn 36 36.75

im chun lp khng chuyn:

NGUYN VNG 3 NGUYN VNG 4

29.5 30.0




S 14.1 S GIO DC V O TO K THI TUYN SINH VO LP 10

AN GIANG TRNG THPT CHUYN THOI NGC HU NM HC 2013 - 2014

CHNH THC

Mn: Ton (chung)

Ngy thi: 15/6/2013


thi ny c 01 trang

Cu 1: (2,0 im)

a) Chng minh rng: 1 1 1

11 2 2 3 3 4

b) Gii h phng trnh: 3x 2y 5 0

2 3x 3 2y 0

Cu 2: (2,0 im)

Cho hai hm s: y = x2 (P) v 1 3

y x2 2

(d)

a) V th ca hai hm s trn cng mt h trc ta . b) Tm ta giao im ca hai th hm s cho.

Cu 3: (2,0 im) Cho phng trnh: x2 + 1(1 y)x + 4 y = 0 (*) a) Tm y sao cho phng trnh (*) n x c mt nghim kp. b) Tm cp s (x; y) dng tha phng trnh (*) sao cho y nh nht.

Cu 4: (4,0 im) Cho tam gic ABC vung cn ti A, D l trung im ca AC, v ng trn (O) ng knh CD ct BC ti E, BD ct ng trn (O) ti F. a) Chng minh rng: T gic ABCF l t gic ni tip.

b) Chng minh rng: AFB ACB v tam gic DEC cn. c) Ko di AF ct ng trn (O) ti H. Chng minh rng: T gic CEDH l hnh vung.






P N CHUNG THI VO LP 10 TRNG THPT CHUYN THOI NGC HU AN GIANG

NM HC 2013 - 2014 Cu 1:

1a) 1 1 1 2 1 3 2 4 3

2 1 3 2 4 31 2 2 3 3 4

2 1 3 2 2 3 1

Vy 1 1 1

11 2 2 3 3 4

.

1b)

13x 2y 5 0

22 3x 3 2y 0

Nhn phng trnh (1) cho 3 ri cng vi phng trnh (2), ta c:

3 3x 3 2y 15 0

2 3x 3 2y 0

5 3x 15 0

5 3x 15

15x 5.

5 3

Thay x 3 vo phng trnh (1), ta c:

3. 3 2y 5 0

2y 2 0

2y 2

2

Vy h phng trnh c mt nghim l 3; 2 . Cu 2:

2a) V th hm s (P) v (d).

Bng gi tr ca hm s (d): 1 3

y x2 2

.

x 0 1

y 3

2 1

Bng gi tr ca hm s (P): y = f(x) = x2.

x -1 0 1

y 1 0 1

2b) Phng trnh honh giao im gia (P) v (d) l:

2

2

1 3 x x

2 2

1 3x x 0

2 2

Gii ra, tta c: 1 23

x 1; x .2




Khi 1 1 1

1 3x 1 y x 1

2 2

Khi 2 2 2

3 1 3 9x y x

2 2 2 4 .

Vy giao im ca hai th l 3 9

1; 1 , ; 2 4

Cu 3:

3a) x2 + (1 - y)x + 4 - y = 0

= (1 - y)2 - 4(4 - y) = 1 - 2y + y2 - 16 + 4y = y2 + 2y - 15.

Phng trnh c nghim kp khi = 0. Khi , ta c: y2 + 2y - 15 = 0

' = 1 + 15 = 16.

y1 = 3; y2 = -5. Vy khi y = 3 hoc y = - 5 th phng trnh c nghim kp. 3b)

x2 + (1 - y)x + 4 - y = 0

x2 + x - xy + 4 - y = 0

x2 + x + 4 - (x + 1)y = 0 Do x, y dng nn x + 1 > 0.

2x x 4y

x 1

4 4y x x 1 1

x 1 x 1

Ta c: 2 2

24 2 2x 1 x 1 4 4 x 1 4

x 1 x 1 x 1

Suy ra: y 4 - 1 = 3. Gi tr ln nht ca y l 3.

Du "=" xy ra khi x + 1 = 2 x = 1 v y = 3. Vy cp s (x; y) tha mn bi l (1; 3). Cu 4:

H

O

F

E

D

CB

A

4a)

0BAC 90 (gi thit) 0CFD 90 (gc ni tip chn na ng trn) T gic ABCF ni tip do A v F cng nhn on BC mt gc bng 900. Vy t gic ABCF ni tip. 4b)




Xt ng trn ngoi tip t gic ABCF. AFB l gc ni tip chn AB ACB l gc ni tip chn AB

Vy AFB = ACB .

Ta c: 0DEC 90 (gc ni tip chn na ng trn) 0DCE 45 (tam gic ABC vung cn) Vy DEC vung cn.

4c) s 1 1AFD DF FH DH2 2

s s s

s 1

DCH DH2

s (gc ni tip)

M AFB ACB

Vy 0DCH ACB 45 . Ta li c tam gic DHC vung nn hai tam gic DEC v DCH vung cn. T gic CEDH l hnh vung.

----- HT -----




S 14.2 S GIO DC V O TO K THI TUYN SINH VO LP 10

AN GIANG TRNG THPT CHUYN THOI NGC HU NM HC 2013 - 2014

CHNH THC

Mn: Ton (chuyn)

Ngy thi: 15/6/2013


Cu 1: (3,0 im)

a) Chng minh rng: 2 3 2 3 2 .

b) Chng minh rng nu a + b + 5c = 0 th phng trnh bc hai ax2 + bx + c = 0, (a 0) lun c hai nghim phn bit.

c) Gii phng trnh: 3x 10x x 16 0

Cu 2: (2,0 im) Cho hm s: y = 2|x| - 1. a) V th hm s cho. b) Tnh din tch tam gic to bi th hm s v trc honh.

Cu 3: (2,0 im)

Cho h phng trnh: 2x y 2 m

3x 4y 8 7m

(m l s cho trc)

a) Gii h phng trnh. b) Tm m h phng trnh c nghim (x; y) sao cho x4 + y4 nh nht.

Cu 4: (3,0 im) Cho hnh vung ABCD ni tip trong ng trn (O); M l im bt k trn cung nh CD; MB ct AC ti E.

a) Chng minh rng gc 0ODM BEC 180 . b) Chng minh rng hai tam gic MAB v MEC ng dng. T suy ra: MC.AB = MB.EC.

c) Chng minh: MA + MC = MB. 2 .






P N MN TON CHUYN THI VO LP 10 TRNG THPT CHUYN THOI NGC HU

NM HC 2013 - 2014 Cu 1:

1a) 2 3 2 3 2

Ta c:

2 2

2 3 2 3 2 4 2 3 4 2 3 3 1 3 1 3 1 3 1 2

Suy ra: 2 3 2 3 2 .

1b) Phng trnh bc hai: ax2 + bx + c = 0.

Do a + b + 5c = 0 b = - a - 5c.

Xt: = b2 - 4ac = (a + 5c)2 - 4ac = a

2 + 10ac + 25c

2 - 4ac = a

2 + 6ac + 9c

2 + 16c

2

= (a + 3c)2 + 16c

2 0.

Du bng xy ra khi:

a 3c 0a c 0

c 0

iu nu khng xy ra do a 0 hay > 0. Vy phng trnh cho lun c hai nghim phn bit.

1c) 3x 10x x 16 0

t: t x x , iu kin: t 0 t2 = x

3 phng trnh tr thnh:

t2 - 10t + 16 = 0

' = 25 - 16 = 9. Phng trnh c hai nghim t1 = 8; t2 = 2.

Vi t1 = 8 3x x 8 x 64 x 4

Vi t2 = 2 3 3x x 2 x 4 x 4

Vy tp nghim ca phng trnh l 3S 4; 4 Cu 2:

2a) y = 2|x| - 1 = 2x 1 x 0

2x 1 x < 0

nu

nu

Vi x 0 th hm s l ng thng y = 2x - 1 qua hai im (0; -1), (1; 1).

Vi x < 0 th hm s l ng thng y = -2x - 1 qua hai im (-1; -3); 1

; 02

V th:

2b) th ct trc honh ti hai im 1

A ; 02

v

1B ; 0

2

th ct Oy ti C(0; -1). Da vo th ta thy tam gic ABC cn ti C c ng cao OC v OC = 1; AB = 1.

Vy din tch tam gic ABC1 1

S OC.AB2 2

vdt.

Cu 3:

3a)

12x y 2 m

23x 4y 8 7m

Nhn phng trnh (1) cho 4 ri cng vi phng trnh (2), ta c:




8x 4y 8 4m

3x 4y 8 7m

11x = 11m x = m. Thay x vo phng trnh (1), ta c: 2m + y = 2 + m

y = 2 - m. Vy h phng trnh c mt nghim (m; 2 - m). 3b)

x4 + y

4 = m

4 + (2 - m)

4

= (m2)

2 - 2m

2(2 - m)

2 + (2 - m)

4 + 2m

2(2 - m)

2

= [m2 - (2 - m)

2]

2 + 2[m(2 - m)]

2

= [4m - 4]2 + 2[2m - m

2]

2

= 16[m - 1]2 + 2[(m - 1)

2 - 1]

2

= 2(m - 1)4 + 12(m - 1)

2 + 2 2.

x4 + y

4 t gi tr nh nht bng 2 khi m = 1.

Vy m = 1 th h phng trnh c nghim l (1; 1) tha mn bi. Cu 4:

O

E

M

D C

BA

4a) Ta c:

OD AC (ng cho hnh vung)

DM MB (gc ni tip chn na ng trn) Vy t gic ODME ni tip.

Suy ra: 0 0ODM OEM 180 ODM BEC 180 1b)

AMB BMC (gc ni tip chn hai cung tng ng AB BC ) ABM ACM (gc ni tip cng chn cung) MAB MEC.

T , suy ra: MA MB AB

MC.AB MB.ECME MC EC

(1)

4c) Ta c:

AMB BMC (gc ni tip chn hai cung bng nhau) MAC MBC (gc ni tip cng chn cung) Vy MAE MBC.

Suy ra: MA AE ME

MA.BC MB.AEMB BC MC

(2)

Cng (1) v (2), ta c: MC.AB + MA.BC = MB.AC

ACAC 2.AB MA MC AB.

AB .

Do AC l ng cho ca hnh vung nn AC 2.AB .




Vy MA MC MB. 2 ----- HT -----





AN GIANG TRNG THPT CHUYN TH KHOA NGHA NM HC 2013 - 2014

CHNH THC

thi chuyn Th Khoa Ngha l thi chuyn Thoi Ngc Hu nm hc 2013 - 2014.





HI PHNG TRNG THPT CHUYN TRN PH NM HC 2013 - 2014

CHNH THC

Mn: Ton


thi ny c 01 trang

Cu 1: (2,0 im) a) Cho biu thc:

x x 3 7 x 10 x 7

A :x 2 x 2 x 4 x x 8 x 2 x 4

Tm x sao cho A < 2.

b) Tm m sao cho phng trnh: x2 - (2m + 4)x + 3m + 2 = 0 c hai nghim phn bit x1, x2 tha mn x2 = 2x1 + 3

Cu 2: (2,0 im)

a) Gii phng trnh: x 7

5x 1 3x 133

b) Gii h phng trnh:

2 2

2 2

2x xy y 3y 2

x y 3

Cu 3: (3,0 im) Cho hai im A, B c nh. Mt im C khc B di chuyn trn (O) ng knh AB sao cho

AC>BC. Tip tuyn ti C ca (O) ct tip tuyn ti A D, ct AB E. H AH CD ti H. a) Chng minh: AD.CE = CH.DE b) Chng minh: OD.BC l hng s. c) Gi s ng thng i qua E vung gc AB ct AC, BD ln lt ti F, G. Gi I l trung im ca AE. Chng minh trc tm G l im c nh.

Cu 4: (1,0 im)

a) Chng minh x y 1 th 1 1

x yx y

.

b) Cho 1 a, b, c 2. Chng minh: 1 1 1

a b c 10a b c

Cu 5: (2,0 im) a) CHo a, b, l 2 s nguyn dng tha mn a + 20; b + 13 cng chia ht 21. Tm s d ca php chia A = 4

a + 9

b + a + b cho 21.

b) C th ph kin bng 20 x 13 vung bng cc ming lt c mt trong hai dng (c th xoay v s dng ng thi c hai dng ming lt) sao cho cc ming lt khng chm ln nhau?







NNG TRNG THPT CHUYN L QU N NM HC 2013 - 2014

CHNH THC

Mn: Ton

(Dnh cho hc sinh thi chuyn ton) Thi gian lm bi: 150 pht. Khng k thi gian giao

thi ny c 01 trang

Cu 1: (2,5 im) a) Tm cc nghim ca phng trnh: 2x2 + 4x + 3a = 0, (1) bit rng phng trnh (1) c mt nghim l s i ca mt nghim no ca phng trnh: 2x2 - 4x - 3a = 0. b) Cho h thc: x2 + (x2 + 2)y + 6x + 9 = 0, vi x, y l cc s thc. Tm gi tr nh nht ca y. Cu 2: (2,5 im)

a) Gii h phng trnh: 4 4

33

x 1 y 1 4xy

x 1 y 1 1 x

b) Tm cc s nguyn x, y sao cho: 2x - 2 y 2 2 2x 1 y .

Cu 3: (3,5 im) Cho on thng BC c M l trung im. Gi H l mt im ca on thng BM (H khc cc

im B v M). Trn ng thng vung gc vi BC ti H ly im A sao cho BAH MAC . ng trn tm A bn knh AB ct on thng BC ti im th hai D v ct on thng AC ti E. Gi P l giao im ca AM v EB. a) t AB = r. Tnh tch: DH.AM theo r. b) Gi h1, h2, h3 ln lt l khong cch t im P n cc ng thng BC, CA, AB. Chng

minh rng: 32 1hh 2h

1AB AC BC

c) Gi Q l giao im th hai ca hai ng trn ngoi tip hai tam gic APE v BPM. Chng minh rng t gic BCEQ l t gic ni tip. Cu 4: (1,5 im) Cho mt thp s (gm 20 vung ging nhau) nh hnh v. Mi vung c ghi mt s nguyn dng n vi 1 n 20, hai vung bt k khng c ghi cng mt s. Ta quy nh trong thp s ny 2 vung k nhau l 2 vung c chung cnh. Hi c th c cch ghi no tha mn iu kin: Chn 1 vung bt k (khc vi cc vung c t tn a, b, c, d, e, f, g, h nh hnh v) th tng ca s c ghi trong v cc s c ghi trong 3 vung k vi n chia ht cho 4?

a e

b f

c g

d h






P N MN TON THI VO LP 10 TRNG CHUYN L QU N NNG

NM HC 2013 - 2014 Cu 1:

a) Cng hai phng trnh vi nhau ta c: (2x

2 + 4x + 3a) + (2x

2 - 4x - 3a) = 0

4x2 = 0 x = 0.

Vi x = 0 th phng trnh (1) tr thnh 3a = 0 a = 0.

Xt a = 0 (1) 2x2 + 4x = 0 x 0

x 2

Tin hnh kim tra vi phng trnh 2x2 - 4x - 3a = 0. V phng trnh (1) c mt nghim x = -2 l nghim i so vi nghim ca phng trnh: 2x

2 - 4x - 3a = 0

Nn vi a = 0, x = 2 lun tha mn. Vy cc nghim ca phng trnh (1) l x = 0, x = - 2. b) Bin i: x2 + (x2 + 2)y + 6x + 9 = 0 thnh x2(1 + y) + 6x + 2y + 9 = 0. (1)

Xt y = - 1 phng trnh (1) tr thnh: 6x + 7 = 0 7

x6

.

Xt y - 1. Ta c: = 36 - 4(1 + y)(2y + 9) = - 8y2 - 44y. phng trnh trn c nghim th 0.

Hay 8y2 + 44y 0 2y2 + 11y 0

11y 0

2 .

Suy ra: Gi tr nh nht ca y l 11

2 t c

2x

3 .

Cu 2:

a) Gii h phng trnh: 4 4

33

x 1 y 1 4xy

x 1 y 1 1 x

iu kin: y 1.

4xy = (x4 + 1)(y

4 + 1) 4x2y2 0 xy 1.

M y 1 x 1.

Do : 33 x 1 y 1 0 1 x

Vy h cho c nghim duy nht (x, y) = (1, 1). b) Ta c:

2x - 2 y 2 2 2x 1 y

2x + y = 2 2x 1 y 2

V x, y nguyn nn 2x 1 y 2 nguyn hay 2x 1 v y 2 nguyn. Ta c phng trnh tng ng:

2 2

2x 1 1 y 2 1 5

V 2x 1 v y 2 nguyn nn




2

2

2

2

2x 1 1 132x 1 1 1 2x 1 2 x2y 2 1 4 y 2 1 2 y 2 3

y 7

2x 1 1 2 2x 1 32x 1 1 4 x 4

y 2 1 1 y 2 2 y 2y 2 1 1

Vy cc s (x, y) tha mn l 3

, 72

v 4, 2 .

Cu 3:

a) Ta pht biu mt b quen thuc:

Cho ABC ni tip ng trn (O) v c AD l ng cao.

Khi , ta lun c: BAH CAO . p dng vo bi ton.

Ta c: BAH CAO vi O l tm ca ng trn ngoi tip ABC.

Suy ra: CAM CAO (1)

Mt khc, ta c: 0 0ABC 90 , ACB 90 nn O

lun nm trong BAC . M [BC] nn O, M cng pha vi AC. T (1), ta c AM i qua O. M O nm trn trung

trc ca BC nn M O hay 0BAC 90 . D thy: BH = HD, AM = MB

DH.AM = BH.MB = 2

21 1 rBH.BC AB2 2 2

.

b)

32 1

PAC PAB 1

ABC PBC 1

1 1

hh 2h 1

AB AC BC

2S 2S 2h1

AB.AC AB.AC BC

2 S S 2h1

AH.BC BC

h 2h1 1

AH BC

1 2

AH BC

Bt ng thc cui cng lun ng v 1 1 2

H BM AH AM BC .2 AH BC

Ta c iu phi chng minh.

c) QBP QMP v QAP QEP nn QBE QMA (g.g)

Do : QB QM

QE QA (2)

v BQE MQA (3)

Q

D

P E

H

A

M CB




T (3), ta c: BQM AQE , kt hp vi (2) th QBM QEA (c.g.c)

Suy ra: QBM QEA (iu phi chng minh) Cu 4:

a e

b x1 x2 f

c x3 x4 x5 x6 g

d h

Ta nh du cc nh trn hnh v.

y cc : ix , i=1,6 u c 4 xung quanh.

Xt theo v tr xi, theo bi, ta c:

1 4

1 2 4

1 2 4

1 2

1 4 2

2 4

4 | x a b x 1

4 | x b x x 2

34 | x a x x

44 | x a b x

4 | 4x 3 a b x x

4 | 3 a b x x

2 44 | a b x x 5 v (3, 4) = 1.

T (5) v (1), (2), (3), (4), ta c:

1 2

1

1

1 4

4 | x x

4 | x a

4 | x b

4 | x x

Vy x1, a, b, x2, x4 ng d (mod4) Lm tng t i vi cc x2, x3, x4, x5, x6. Vy ta c t nht 12 s ng d (mod4). M: T 1 n 20 ch c 4 lp s, mi lp gm 5 s ng d (mod4) v 12 s ny phi khc nhau. Vy khng c cch xp no tha mn yu cu bi ton.

----- HT -----





CN TH TRNG THPT CHUYN L T TRNG NM HC 2013 - 2014

CHNH THC

Mn: Ton


thi ny c 01 trang

Cu 1: (2,0 im) Cu 2: (1,5 im) Cu 3: (3,5 im) Cu 4: (1,0 im)







YN BI TRNG THPT CHUYN NGUYN TT THNH NM HC 2013 - 2014

CHNH THC

Mn: Ton


thi ny c 01 trang

Cu 1: (1,5 im) Cho biu thc: a 1 a b3 a 3a 1

P :a ab b a a b b a b a ab b

a) Tm iu kin ca a, b P c ngha, ri rt gn P. b) Tm cc gi tr ca a Q = P(3a + 5) nhn gi tr nguyn.

Cu 2: (3,0 im)

1. Gii h phng trnh: 2 2x y xy 3y 4

2x 3y xy 3

2. Cho phng trnh: x2 - mx + 1 = 0 (1) (vi m l tham s) a) Xc nh cc gi tr ca m hai nghim x1, x2 (nu c) ca phng trnh (1) tha mn ng thc: x1 - 2x2 = 1 b) Xc nh cc gi tr ca m phng trnh (1) c hai nghim phn bit u ln hn 2.

Cu 3: (3,5 im) Cho na ng trn (O; R) ng knh AB, ly M l im ty thuc na ng trn (M khng trng vi A v B). K ng cao MH ca tam gic MAB. Gi E v F ln lt l hnh chiu ca H trn MA v MB. a) Chng minh t gic ABFE ni tipng mt ng trn. b) Ko di EF ct cung MA ti P. Chng minh MP2 = MF.MB. T suy ra tam gic MPH cn.

c) Xc nh v tr ca im M trn na ng trn (O) t gic MEHF c din tch ln nht. Tm din tch ca t gic theo R.

Cu 4: (1,0 im) Tm nghim nguyn ca phng trnh: 2x2 + 3y2 + 4x - 19 = 0.

Cu 5: (1,0 im)

Cho ba s dng x, y, x tha mn iu kin 1 1 2

0x y z .

Tm gi tr nh nht ca biu thc: x z z y

T2x z 2y z







THI BNH TRNG THPT CHUYN THI BNH NM HC 2013 - 2014

CHNH THC

Mn: Ton


Cu 1: (2,0 im)

Cho biu thc: x 1 1

P x 4x 4x 2 x 2

vi x 0, x 4.

1. Rt gn biu thc P. 2. Tm gi tr nh ca P.

Cu 2: (2,0 im)

Cho h phng trnh: mx y 1

x my m 6

(m l tham s)

1. Gii h phng trnh vi m = 1. 2. Tm m h s nghim (x; y) tha mn: 3x - y = 1.

Cu 3: (3,5 im) 1. Cho phng trnh bc hai: x2 - (2m - 1)x + m2 - m - 6 = 0. (m l tham s). Chng minh phng trnh lun c hai nghim phn bit x1; x2 vi mi gi tr ca m. Tm m : -5 < x1 < x2 < 5. 2. Gii phng trnh: (x + 2)(x - 3)(x2 + 2x - 24) = 16x2.

Cu 4: (1,0 im) Cho tam gic u ABC c ng cao AH. Trn ng thng BC ly im M nm ngoi on BC sao cho MB > MC v hnh chiu vung gc ca M trn AB l P (P nm gia A v B). K MQ vung gc vi ng thng AC ti Q. 1. Chng minh 4 im A, P, Q, M cng nm trn mt ng trn. Xc nh tm O ca ng trn . 2. Chng minh: BA.BP = BM.BH. 3. Chng minh OH vung gc vi PQ. 4. Chng minh: PQ > AH.

Cu 5: (0,5 im) Gii phng trnh:

33

2 2

2013x 1 2013x 12x 2014 x 2013 x 1

2 x 2 x







NINH BNH TRNG THPT CHUYN LNG VN TY NM HC 2013 - 2014

CHNH THC

Mn: Ton


thi gm c 05 cu 01 trang

Cu 1: (1,5 im)

1) Rt gn biu thc: M 2 2 8 18

2) Gii h phng trnh: 2x y 9

3x 2y 10

Cu 2: (2,0 im)

Cho biu thc: 2

2

2x 4 1 1A

1 x 1 x 1 x

(vi x 0, x 1)

1) Rt gn A. 2) Tm gi tr ln nht ca A.

Cu 3: (2,0 im) Cho phng trnh: x2 - 2(m + 1)x + 2m = 0 (1) vi x l n, m l tham s) 1) Gii phng trnh (1) vi m = 0. 2) Tm m phng trnh (1) c hai nghim l di hai cnh gc vung ca mt tam gic

vung c cnh huyn bng 12 .

Cu 4: (3,0 im) Cho na ng trn tm O ng knh AB. Mt im C c nh thuc on thng AO (C khc A v C khc O). ng thng i qua C v vung gc vi AO ct na ng trn cho ti D. Trn cung BD ly im M (M khc B v M khc D). Tip tuyn ca na ng trn cho ti M ct ng thng CD ti E. Gi F l giao im ca AM v CD. 1) Chng minh t gic BCFM l t gic ni tip. 2) Chng minh: EM = EF. 3) Gi I l tm ng trn ngoi tip tam gic FDM. Chng minh ba im D, I, B thng hng, t suy ra gc ABI c s o khng i khi M di chuyn trn cung BD.

Cu 5: (1,5 im) 1) Chng minh rng phng trnh: (n + 1)x2 + 2x - n(n + 2)(n + 3) = 0 (x l n s, n l tham s) lun c nghim hu t vi mi s nguyn n.

2) Gii phng trnh: 3 25 1 x 2 x 2 .







VNH PHC TRNG THPT CHUYN VNH PHC NM HC 2013 - 2014

CHNH THC

Mn: Ton chung

(Dnh cho hc sinh thi chuyn) Thi gian lm bi: 150 pht. Khng k thi gian giao

thi ny c 01 trang

Cu 1: (2,0 im) Cho biu thc 3x +1

P = - x : x -1x +1

, vi x 1, x 1 .

a) Rt gn biu thc P . b) Tm tt c cc gi tr ca x P = x2 - 7.

Cu 2: (2,0 im).

a) Gii h phng trnh:

2 3- = -1

x y -1

3 1+ = 4

x y -1

b) Gii phng trnh: x +1 x + 2 x +3 x + 4

+ = +99 98 97 96

Cu 3: (2,0 im) Cho phng trnh: x2 - (2m - 1)x + m - 2 = 0, (x l n, m l tham s). a) Gii phng trnh cho vi m = 1. b) Tm tt c cc gi tr ca tham s m phng trnh cho c hai nghim v tng lp phng ca hai nghim bng 27.

Cu 4: (3,0 im) Cho ng trn (O) v im M nm ngoi (O). T im M k hai tip tuyn MA, MC (A, C l cc tip im) ti ng trn (O). T im M k ct tuyn MBD (B nm gia M v D, MBD khng i qua(O). Gi H l giao im ca OM v AC. T C k ng thng song song vi BD ct ng trn (O) ti E (E khc C). Gi K l giao im ca AE v BD. Chng minh: a) T gic OAMC ni tip. b) K l trung im ca BD.

c) AC l phn gic ca gc BHD .

Cu 5: (1,0 im) Cho cc s thc dng a, b, c tha mn a2 + b2 + c2 = 1. Chng minh: 2 2 2

2 2 2

ab 2c bc 2a ca 2b2 ab bc ca

1 ab c 1 bc a 1 ca b






P N MN TON CHUNG THI VO LP 10 TRNG THPT CHUYN VNH PHC

NM HC 2013 - 2014 Cu 1:

1a)

Rt gn biu thc:

2x 1 x x 1

P x : x 1x 1

= 2= x - 2x +1 : x -1 = x - 1.

Vy P = x- 1.

1b) Theo phn a) ta c 2 2P x 7 x 1 x 7 1

2x 2

1 x x 6 0x 3

.

Kt lun cc gi tr ca x cn tm l: x = -2

x = 3

Cu 2:

2a) iu kin xc nh: x 0, y 1 . t 1 1

a , bx y 1

Thay vo h cho ta c

2a 3b 1 2a 3b 1 11a 11 a 1

3a b 4 9a 3b 12 2a 3b 1 b 1

x 1 x 1

y 1 1 y 2

.

Vy h phng trnh cho c nghim l (x; y) = (1; 2).

2b) Gii phng trnh: x 1 x 2 x 3 x 4

99 98 97 96

rng 99 1 98 2 97 3 96 4 nn phng trnh c vit li v dng

x 1 x 2 x 3 x 41 1 1 1

99 98 97 96

(1)

Phng trnh (1) tng ng vi

x 100 x 100 x 100 x 100 1 1 1 1

x 100 0 x 10099 98 97 96 99 98 97 96

.

Vy phng trnh cho c nghim duy nht x = -100.

Cu 3:

3a) Khi m = 1 phng trnh c dng 2x x 1 0 .

Phng trnh ny c bit thc 2( 1) 4.1.( 1) 5 0, 5

Phng trnh c hai nghim phn bit 11 5

x2

v 2

1 5x

2

3b) Phng trnh cho c bit thc:

2 2 2(2m 1) 4.1.(m 2) 4m 8m 9 4(m 1) 5 m, 0

Vy phng trnh c hai nghim phn bit 1 2x , x vi mi gi tr ca tham s m.

Khi , theo nh l Vit: 1 2 1 2x x 2m 1, x x m 2

Ta c 3 3 3 3 21 2 1 2 1 2 1 2x x (x x ) 3x x (x x ) 8m 18m 21m 7




3 3 3 2 21 2x x 27 8m 18m 21m 34 0 (m 2)(8m 2m 17) 0 (1)

Do phng trnh 28m 2m 17 0 c bit thc 4 4.8.17 0 nn (1) m 2

Vy m = 2.

Cu 4:

4a) T gic OAMC ni tip.

Do MA, MC l tip tuyn ca (O) nn 0OA MA, OC MC OAM OCM 90 0OAM OCM 180 T gic OAMC ni tip ng trn ng knh OM.

4b) K l trung im ca BD.

Do CE//BD nn AKM AEC , AEC ACM (cng chn cung AC ) AKM ACM . Suy ra t gic AKCM ni tip.

Suy ra 5 im M, A, K, O, C cng thuc ng trn ng knh OM 0OKM 90 hay OK vung gc vi BD. Suy ra K l trung im ca BD.

4c) AH l phn gic ca gc BHD . Ta c: 2MH.MO MA , 2MA MB.MD (Do MBA, MAD ng dng) MH.MO MB.MD

MBH, MOD ng dng BHM ODM t gic BHOD ni tip MHB BDO (1)

Tam gic OBD cn ti O nn BDO OBD (2)

T gic BHOD ni tip nn OBD OHD (3)

T (1), (2) v (3) suy ra MHB OHD BHA DHA AC l phn gic ca gc BHD . Cu 5:

Do 2 2 2a b c 1 nn ta c

2 2 2 2

2 2 2 2 2 2 22 2 2

ab 2c ab 2c ab 2c ab 2c

1 ab c a b c ab c a b ab ab 2c a b ab

p dng bt ng thc x y

xy , x, y 02

2 2 22 2 2

2 2 2 2 2 22 a b c2c a b 2ab

ab 2c a b ab a b c2 2

2 2 22

2 2 2 22 2 2

ab 2c ab 2c ab 2cab 2c 1

1 ab c a b cab 2c a b ab

B

K

E

H

C

A

O M

D




Tng t 2

2

2

bc 2abc 2a

1 bc a

(2)

v 2

2

2

ca 2bca 2b

1 ca b

(3)

Cng v theo v cc bt ng thc (1), (2), (3) kt hp 2 2 2a b c 1 ta c bt ng thc cn chng

minh. Du = khi 1

a b c3

.

---- HT ----




S GIO DC V O TO K THI TUYN SINH VO LP 10 VNH PHC TRNG THPT CHUYN VNH PHC

NM HC 2013 - 2014 CHNH THC

Mn: Ton chuyn


thi ny c 01 trang Cu 1: (3,0 im).

a) Gii h phng trnh:

xy x y 1

yz y z 5 x, y, z

zx z x 2

b) Gii phng trnh: 2 2x 3x 2 x 1 6 3 x 1 2 x 2 2 x 1, x .

Cu 2: (2,0 im).

a) Chng minh: Nu n l s nguyn dng th 2013 2013 20132 1 2 ... n chia ht cho n(n+1). b) Tm tt c cc s nguyn t p, q tha mn iu kin 2 2p 2q 1 .

Cu 3: (1,0 im) Cho a, b, c l cc s thc dng tha mn abc = 1. Chng minh:

a b c 3

a 1 b 1 b 1 c 1 c 1 a 1 4

Cu 4: (3,0 im)

Cho tam gic nhn ABC, AB < AC. Gi D, E, F ln lt l chn ng cao k t A, B, C. Gi

P l giao im ca ng thng BC v EF. ng thng qua D song song vi EF ln lt ct

cc ng thng AB, AC, CF ti Q, R, S. Chng minh:

a) T gic BQCR ni tip.

b) PB DB

PC DC v D l trung im ca QS.

c) ng trn ngoi tip tam gic PQR i qua trung im ca BC.

Cu 5: (1,0 im)

Hi c hay khng 16 s t nhin, mi s c ba ch s c to thnh t ba ch s a, b, c tha

mn hai s bt k trong chng khng c cng s d khi chia cho 16?






P N MN TON CHUYN THI VO LP 10 TRNG THPT CHUYN VNH PHC

NM HC 2013 - 2014 Cu 1:

1a)

x 1 y 1 2xy x y 1

yz y z 5 y 1 z 1 6

zx z x 2 z 1 x 1 3

Nhn tng v cc phng trnh ca h trn ta c

2 x 1 y 1 z 1 6x 1 y 1 z 1 36

x 1 y 1 z 1 6

Nu x 1 y 1 z 1 6 , kt hp vi h trn ta c

x 1 1 x 2

y 1 2 y 3

z 1 3 z 4

Nu x 1 y 1 z 1 6 , kt hp vi h trn ta c

x 1 1 x 0

y 1 2 y 1

z 1 3 z 2

.

Vy h phng trnh cho c 2 nghim x; y; z 2; 3; 4 , 0; 1; 2 . 1b) iu kin xc nh x 1 . Khi ta c

2 2x 3x 2 x 1 6 3 x 1 2 x 2 2 x 1

x 1 x 2 x 1 x 1 6 3 x 1 2 x 2 2 x 1

x 1 x 2 x 1 x 1 3 x 1 2 x 1 2 x 2 6

x 1 x 2 x 1 3 2 x 1 x 2 3

x 1 2 x 2 x 1 3 0

2x 2 x 1 3 0 x 2 x 1 2 x 2 x 1 9 x x 2 4 x

2 2

x 4x 2

x x 2 x 8x 16

x 1 2 x 1 4 x 3.

Vy phng trnh cho c tp nghim l S = {2; 3}.

Cu 2:

2a) Nhn xt. Nu a, b l hai s nguyn dng th 2013 2013a b a b .

Khi ta c: 20132013 2013 2013 2013 2013 2013 2013 20132 1 2 ... n 1 n 2 n 1 ... n 1 n 1

(1)

Mt khc

2013 2013 2013

2013 2013 20132013 2013 2013 2013

2 1 2 ... n

1 n 1 2 n 2 ... n 1 1 2.n n 2

Do n(n + 1) = 1 v kt hp vi (1), (2) ta c 2013 2013 20132 1 2 ... n chi

Documents

[VNMATH.com] 77 de Thi Vao Cac Lop 10 Truong Chuyen 2013