8/6/2019 THI TUYN SINH LP 10 NM 2011

1/5

1 Ths. Nguyn Tng V Trng Ph Thng Nng Khiu

HNG DN GII THI TUYN SINH LP 10 NM 2011 Mn thi: TON (chuyn)

Thi gian lm bi: 150 pht, khng th thi gian pht .Cu I.Cho phng trnh ( + 3) + = 0, trong l tham s saocho phng trnh c hai nghim phn bit

,

a) Khi = 1, chng minh rng ta c h thc + = 2 + 2 + 6b) Tm tt c cc gi tr sao cho + = 5c) Xt a thc () = + + . Tm tt c cc cp s (, ) sao cho

ta c h thc () = ()vi mi gi tr ca tham s mLi gii.Ta c = (m+3)2 4m2 = 3(m+1)(3-m). > 0 -1 < m < 3.

a)

Khi m = 1, phng trnh c hai nghim dng x1, x2. Theo nh l Vit ta cx1 + x2 = 4, x1.x2 = 1. Ta bin i tng ng

626622

626222622

2121214

2121

42

41

821

42

41

82

81

xxxxxxxxxx

xxxxxxxx

H thc cui cng ng do x1 + x2 = 4 v x1x2 = 1.

b) phng trnh c hai nghim khng m th0

0 1 3

0

S m

P

(*)

Theo nh l Viet th x1 + x2 = m+3 v x1x2 = m2

. Ta c.2||2523525 2212121 mmmmxxxxxx

Gii phng trnh ny, ta c hai nghim .2,3

2 mm So snh vi iu

kin (*) trn ta loi nghim m = -2.

Vy gi trm 521 xx l3

2m .

c) Ta c P(x1) = P(x2) x13 x23 + a(x12-x22) + b(x1-x2) = 0

(x1-x2)(x12

+x1x2+x22

+a(x1+x2)+b) = 0 (x1-x2)((x1 +x2)

2 x1x2 + a(x1+x2)+b) = 0

(x1-x2)((m+3)2

m2

+ a(m+3)+b) = 0

(6+a)m + 9 + 3a + b = 0 vi mi - 1 < m < 3

a = - 6, b = 9.

Vy (a, b) = (-6, 9) l cp s cn tm.

8/6/2019 THI TUYN SINH LP 10 NM 2011

2/5

2 Ths. Nguyn Tng V Trng Ph Thng Nng Khiu

Cu II.a) Cho , l cc s thc dng. Tm gi tr nh nht ca biu thc

= 1 + 1 + 1 +

b) Cho x, y , z l cc s thc tha mn iu kin || 1, || 1, || 1.Chng minh rng ta c bt ng thc:1 + 1 + 1 9 ( + + )

Li gii.a) Ta c + 2, suy ra 1 + + + 1 + 2 +

(1 + )(1 + ) (1 + ) 1 + . 1 + 1 + (v 1+ ab > 0) 1 (v 1 + ab > 0)Du = xy ra khi v ch khi a = b.Vy minP = 1 khi v ch khi a = b.

b) Bnh phng hai v ca bt ng thc, ta c bt ng thc tng ng2222222222 )(9112112112111 zyxxzzyyxzyx

zxyzxyxzzyyx 111111111 222222 .

hon tt php chng minh, ta ch cn chng minh xyyx 111 22 . Tht

vy, do 1 xy 0 nn (*) tng ng vi

(1 x2

)(1 y2

) (1 xy)2

(x y)2

0 (hin nhin ng)

Cu III. Cho tam gic ABC c AB = b, AC = b. M l mt im thay i trn cnhAB. ng trn ngoi tip tam gic BMC ct AC ti N.

a) Chng minh rng tam gic AMN ng dng vi tam gic ACB. Tnh t s din tch tam gic AMN bng mt na din tch tam gic ACB.

b) Gi I l tm ng trn ngoi tip tam gic AMN. Chng minh rng I lunthuc mt ng thng c nh.

c) Gi J l tm ng trn ngoi tip tam gic BMC. Chng minh rng diIJ khng i.Li gii.

8/6/2019 THI TUYN SINH LP 10 NM 2011

3/5

3 Ths. Nguyn Tng V Trng Ph Thng Nng Khiu

a) Theo tnh cht ca t gic ni tip ta c ANM = MBC = ABC. Mtkhc NAM = BAC. Suy ra hai tam gic AMN v ACB ng dng.

Thai tam gic ng dng ny ta suy ra AM.AB = AN.AC. din tch tamgic AMN bng mt na din tch tam gic ACB th t sng dng phi bng

2

1, tc l

2

1

AC

AM. Suy ra

2

cAM . Ty ta tnh c

2

cbBM . Suy

ra .2 cb

c

BM

AM

b) Cch 1. Gi O l tm ng trn ngoi tip tam gic ABC v AH l ngcao ca tam gic ABC. Khi d thy rng OAC = BAH. T , doANM = ACH nn ty ta suy ra OA vung gc MN, suy ra AO lng cao trong tam gic ANM. Ty, cng do NAO = MAH.

Mt khc do tam gic AMN v ABC ng dng nn suy ra IAM v OAC ng

dng, do IAM = CAO = MAH, nn tm I ng trn ngoi tip tamgic AMN nm trn AH cnh (pcm). (1 )

Cch 2 chng minh AO NM: V tip tuyn Ax ca (O), ta c OA Ax.xAM = ACB =AMN, suy ra Ax//MN.Do OA MNCch 2

Ta c I l tm ng trn nn AIM = 2ANM = 2ABC.

M tam gic AIM cn ti I nn MAI = (1800- AIM) = 900 - ABC.

8/6/2019 THI TUYN SINH LP 10 NM 2011

4/5

4 Ths. Nguyn Tng V Trng Ph Thng Nng Khiu

Suy ra MAI + ABC = 90 nn AI BC, I thuc ng cao AH ca tam gicABC.

c) V hai ng trn ngoi tip hai tam gic ABC v BMC c chung dy cungBC nn OJ vung gc vi BC. Theo chng minh trn th AI vung gc viBC. Suy ra AI // OJ. Tng t, IJ vung gc vi MN v AO vung gc viMN theo cu a), suy ra IJ // AO. Suy ra AIJO l hnh bnh hnh. Suy ra IJ =

AO = R khng i (pcm).

Cu IV. Cho , , l cc s nguyn sao cho 2 + , 2 + , 2 + u l cc schnh phng (*).

a) Bit rng c t nht mt trong ba s chnh phng ni trn chia ht cho3. Chng minh rng tch ( )( )( )chia ht cho 27.

b) Tn ti hay khng cc s nguyn , , tha iu kin (*) sao cho( )( )( )chia ht cho 27?

Li gii.a) Gi s 2a + b = m2, 2b + c = n2, 2c + a = p2.

Cng ba ng thc li, ta c 3(a + b + c) = m2 + n2 + p2. Suy ra m2 + n2 + p2 chiaht cho 3.Ch rng bnh phng ca mt snguyn chia 3 d 0 hoc 1. Do nu c 1trong 3 s, chng hn m chia ht cho 3 th n2 + p2 chia ht cho 3 v nh th n2 vp

2cng phi chia ht cho 3.

Cui cng, ch rng nu 2a + b chia ht cho 3 th a b = 3a (2a+b) chia ht cho3. Tng t ta c b c v c a chia ht cho 3, suy ra (a b)(b c)(c a) chia ht

cho 27.

b) Tn ti. Chng hn c th ly a = 2, b = 0, c = 1.Cu V. Cho hnh ch nht ABCD c AB = 3, BC = 4.

a) Chng minh rng t 7 im bt k trong hnh ch nht ABCD lun tmc hai im m khong cch gia chng khng ln hn 5b) Chng minh rng khng nh cu a) vn cn ng vi 6 im bt knm trong hnh ch nht ABCD.

Li gii.a) Chia hnh ch nht 3 x 4 thnh 6 hnh ch nht con 1 x 2. Theo nguyn l

Dirichlet, tn ti 2 trong 7 im cho thuc vo 1 hnh ch nht v dong knh ca hnh ch nht ny bng 5 nn ta c iu phi chng minh.

8/6/2019 THI TUYN SINH LP 10 NM 2011

5/5

5 Ths. Nguyn Tng V Trng Ph Thng Nng Khiu

b) Chia hnh ch nht thnh 5 phn nh hnh v.

Nhn xt v thi.- Cc cu d l 1ab, 3a ( ng dng).- Cc cu m c trung bnh l 2ab, 4b, 5a, 4b, 1c- Cc cu kh (phn loi hsg) l 4a, 3c, 5b.