Bộ đề thi vào lớp 10 bộ môn toán trường chuyên năm 2017

NcuvEN . THI ruyrN sINH vAo Lop 10 THpr
oAo r4o
NAu HoC: 2Ot7 - 2018
ur6N THr: roAN Hec
Bn cuixH THrIc
' (Dnnh cho thi sinh thi viro chuy6n

:r
,:

UBND TINH rHAt
sd crAo DUC vA

crruytx

To6n)

Thdri gian

lim biri:

phft

180

cau 1(1,0 ili6m). Kh6ng drrng m6y tinh cam tay h6y rut

(kh1ng ke thdi gian giao

A=

so.n

di)

ffl';*)

Ciu2(2,0 di6m). ciei he phuong trinh sau:

I*'*y'-xy+4y+l=0

Ir(r- *'- y'*r*)=z(*' +t)'
Cflu 3(1,0 di6m). Cho s0 tg nhi6n A =W
- l8 + 2n vli n e N,z ) 2.
n chilsdT

Chtmg minh ring
CAu 4(1,5

A chia h6t cho 9.

ili6m). Cho a,b,c ld c6c sti duong th6a m6n

Tim giritrilonnh6t criabi6uthtc: P

Ji

a+b+cs

-+*L+*.
Jo'+t Jo'+t

cffu 5(1,0 ili6m). vdi m6i sti nguy€n ducing n taky hiQu a,

li

si5

Jc2

.

+t

nguyEn g6n

Ji *at.

Vf dU: o, =l;a, =l;az = 2;aq - 2;as = 2;ao = 2;o, = 3 . Tinh gi6 tri cria t6ng:

1*
^s=l*f*l*...*
at a2 a3
dzon
CAu 6(1,0 ili6m). Cho hai duong trdn t6m
duong trdn

(Q)

duong trdn tdm

ta,

O

ei6*

l.

Phan k6o ddi

I
azots

Q vd O, nimngodi

nhau. DoAn tha"g

eO, c*t

cta cloqn OrQ cdtdudrng trOn (q) @i B. Drmg

ti€p xric ngodi vdi dudmg trdn (Q

(Or) tqi C (tli6m O khdng nim tren do4n QOr).

) tai O vd titip xric trong v6i duong trdn

Chrmg minh rdng c6c dii:m

A,B,C,D cing

nim tr6n mQt duong trdn.

Ciu 7Q,5 tliGm). Cho ba AB vir

BCFK. Gqi

/

ld trung cti6m cria do4n

thing

citcacducrngthdng BD vd AB ]ffilnlugtt4i
a.

M

Dutng thlng qua ^I vu6ng gbc vfii EF
vd ff. chungminhring:
EF, .

Tri giirc AEIN vi tri gi6c EMID nQi titip dugc trong dudng trdn.

b. Ba dii:m

A,I,D

thang hang vd c6c cti6m

c. Ba duong thdng AK, EF

,CD


quy.

B,N,F,M,E nim tr6n cirng mQt dudng trdn.


rgar NcuypN
so crAo oqc vl oao r4.o
UBND riNH

HITOI{G oAN CHAM
THI TUYTX STNTT LOP 10 THPT CHUYTN
xAna Hec zotz.- 2ot8
wION THr: ToAN IreC
(Dlrnh cho thf sinh thi vlro chuy6n To6n)

( Bdn hudng ddn chiim

gim c6 06 trang)

l'#t#?f,*"Xx?fviing

ycu cAu cria hu6ng dan ch6m a6 arenn gi6 irung bdi

ldm cria thi sinh. Thf sinh ldm c6ch kh6c tl6p 6n niSu dring vin cho tli6m t6i tta.
- Khi v4n dpng ddp invi thang dir5m, gi6m kh6o c6n chri dQng, linh ho4t vdi tinh
thAn trdn trgng bdi lim cria hgc sinh.
- Ni5u c6 viQc chi titit h6a cti6m chc y cdn phai d6m bio kh6ng sai lQch v6i t6ng
"
dii5m vd dugc th6ng nh6t trong todn hQi d6ng ch6m thi.
- Dii6m toan bii li t6ng di6m cria c6c cdu h6i trong de thi, ch6m
vi

kh6ng ldm trdn.

II. Eip 6nn vir thans di6m
Ciu
Cf,u

1

1r0

N6i dune

Di6m

Ta c6:

(r.G)

.---

Jro + Jz

_.6-G(:...6)

0r25

o(6.t)

.fG(:.G)

0.25

z(.,6 + r)

_(S-,Xr.6)
=@(r.r)
-z(S+r)
z(Js

3'',6+5-3-\6
Cdu2
2r0

+2
zJs +z

2"15
-:-l

+

t)

_l

, l*'+y2 -ry+4y+l=0
ra co
b(, - x' - y' +zry) -z(x' +t)
'
+xy- y
l*'*l=-f2
*
tr(, - x' - y' *z'y) = z(-y' * *y l*'*l=-!2 +ry-4y

*

0r25

0.2s
t

ay)

0r25

tr[,5-(, - y)' -z(x-r)]= o(z)

Ta thdv v

0 khdns ld nshiEm cria h6.

/n

at


a

(2)

*

l*'*L=-r2 +xy-4y*
1,5-(, - y)' -z(*- y)=,

ll*-v=-5
I 1r' ., = -y' + xy - +r(3)

I

<+l

ll.-'= )-5
t:;

.i:-y'*xy-4y

,

l[x-v=3

Lto ., = -y' + xy - +r(+)

0.5

(x-v=-5

Giai (3) <+

tr' *, = -y' + xy - 4y
lx= y-5

<+{

L(y

- 5)' * t = -!2 * (y - s)t - at

lx= v-5
v6 nghiQm)
i
*r
o(he
=
tr, ,

o
Giai (4) <+

lx-

v

=3

1r' ; t= -!2 + xy -4y
y+3
e
[(y* 3)' *r=-!2 *(y*
y+3
e{ lx ly"

0.5

+7 y +10

flx=l

3)t-at

-0

0.5

- = -,
el 11,
I {x=-2
ti

L[.Y=-:

vay hg c6 nghiQm

Ciu
1r0

3

(t;-z);(-z;-s)

Cho sti t.u nhi6n a e N *.

DIt S(")

ta t6ng c6c chil sti cria a . Khi d6 ta

0.2s

lu6n c6 a - S (a) chia hi5t cho 9.

Tac6 A=777 J-18+ 2n
zchBsdT

T0ng cdc chtr

(_

siS

ctia

\

lllJ

-nl+9(n-2\
-7ll]-l-18+ 2n-TllllJ
i-| \
'
nch8sdl

111-l ld r

ffiffi

n€n ta c6 111-1

-nlchiahiStchog

Hi6n nhi6 n

9(n- 2) chia hi5t cho 9

)

-n

)

chiah6t cho 9.

Vi

0.25
0.25

vftv 7l L--/
I

\nchfisdt

\zchEsdt

|

Do d6 A:.9

'
0.25

2


Cflu 4

Tac6:
(a + b +

")"
Do d6 ta c6:

3(ab + bc +

ab+bc+ac1l=

a2

Tuong t.u ta c6
(a + o)(b+ r) <

+t;(a

b2

ac)+

3>

3(ab + bc + ac)

+ab+bc+ac1a2
+

+ ab + bc + ac
+l*(a+b)(a+")<

a2 +1

c)(t+ r) < c2 +t

0.25

Ti, d6 ta c6:
a

P<
a+

b)(a+ c)

a+b)(b

a+c)(b+c)

+ c)

0.25

(a+c)(b+ c)
Theo BDT Cauchy ta c6:

t( b b \
2\a+b b+c)'

?-l-r-1.

(a +b)(b + c)

I

c'

0.25

-')

' b+c)
2\a+c
(o+c)(b+")'L(

N€n:

l( a a T-'T- b b T-T7 c c )
-l
Z\a+b a+c a+b b+c a+c b+c)
?
+P.f :

r

2

a=b--c
D6u ding thric

xiy ra khi

o+b+, -rl3

aa
OA=[=g=a+b=-a+c

.s
3

bb

a+b b+c
cc
a+c=-b+c

0.25

,l

,J'

1

Vpy P c6 giiltri lon nh6t bdne
"2: khi vd chi khi Cl=b=C=-.

Ciu

5

Gqi & ld sO nguy6n duong ge"

1r0

1,

J;

nh6t.

rni

d6 ta c6:

Ji < k +L e k' - k +!.n < k2 + t +L
-!.2244

+k2 -k+l
3


0.25

Ttd6sEc6 2k sd ao=k.
v0y
s=

[l*l)*
t)
[l
2

(

L*...*1)*

2)

\2

sdhang

4 sdhang

--

(tir a,rr, = 0ts82 = ... =
S

Cffu 6

=2.44+:g.a45

(khdng

vG

L+...+l)* ( L*...*l)
\44 44) (4s 4s)

...*(

aror,

88 sti

hpg

0.25

38 sdf hang

= 45)

3998

0.2s

45

hinh kh6ng cho tli6m)

1r0

Dilffi=a;@D=f
0.25

'2 -q=@=1800 -2G)g1
M[t khrac: @ -@,= 1800 -(o * p)(**)
Tac6

dE),-ffir=eoo

Tt (*) vA (**)

ta c6

0.25

dB) = " !^0 0)
\/

2

Ta c6

ffi,

-1800 - (m, *ffi)=

Tu(l) vir(2)suyra

0.25
1800

-(ro' - *.e00 - ;)=

ffi=6)-a+B2

Do d6 tir giitc CBDA nQi tiiip dulng trdn.

'

ffe>
0.25


I

;

t-

:

i

(kh6ng vE hinh khdng cho tli6m)
M

Ciu

7

215

a.Tac6:

fu =ffi

= 900 n€n

ffi -ffi

b. Ta c6

Gqi O

li

tf

= 900 n6n

frE

gi6c AEINnQi titip tluong trdn dudrng kfnh

tu

gi6c

Ell

EMIDnQi titip dudrng trdn duln gkinh EM

=fu - 450 n€n ADllBr (t)

giao cli€m ci,r- BE vir AD. Khi d6 O

li

trung rti6m ctr- BE vit AD

Ma 1ld trung cti6m cia EF nln OI ld ducrng trung binh cta tam gi6c EBF
fmg vdi canh BF

Do d6

Tir

(l)

oI

ll

BF (2)

vA (2) ta c6 A, I,

D ctng nim tr6n dulng thdng qua O vd song song

v6i BF
Vdy A, I, D thinghdng.
Ta c6:

Tam gi6c NEF cdn tai.Ail,

fu -ifu

-450n€n tam gi6c NEF vu6ng cdn

fu =m

= 45on6n tam gi6c MEFvuOng cdn

tai N.

Tam gi6c MEF cdnt1i M,

0.25
0.25
0.25


Vi vay tri giSc MENF le hinh vu6ng
Suy ra bi5n cti6m M,

E,lf f ni*

tr€n dudmg trdn dudrn gkinh MN

0.25

M4t kh6c tam gi6c MBN vuOng t4i B n€n .B cflng nim ffen dudmg trdn
iludrng kirh

MN.

VAy 5
B,N,F,M ,E nimtrOn ctng mQt dulng trdn

I
I

I

c. Chmg minh tuong tg ta c6 ba di6m C,

Ggi G

li

giao itii5m ctr,AKvit

ducm gkinh

I, Kthdng hdng

CD.

MN.

I O.ZS
I
I

I

D

li

trgc t6m tam gi6c AKC

Trigi6c AEGDnQititip

Tt

n€,n

frd = 900(3)

I

n€'nfu=ffi,=450(4)

giric CGKF nQi ti6p n€n

GF

t--

=fu

= 450(5)

^

I Ti, (3), (4) va (5) ta co EGA + AGC + CGF = 1800

O.rt

lO.r,
0.25

+

EGF = 1800

N€n G nim tr6n ducrng thtng EF

Vfy

ba duong thing AK, CD, Ef'ddng quy.

-- rudt----

0.25


srng YAo 16r t0 THpr cnuv.&N
xAu Hec: zot6 - zot7
nnoN THI: TO.A,N HQC
@irnh cho thi sinh thi vio chuy0n To{n)
Thli giau lim bii: 150 phrf,t (kh$nS kd thdi gian giaa dA)

uBND T'iNH ruAlNeuyEr.r THI TUyEN
sO cll.o DUC vA uAo

r.So

on cuiNn

rntfc

cflu r (r,s di€m). ciei

Cflu 2

1:t,O

hQ phucrng

trinh:

l'r(*' - !')=3*
* y'r)'_tty
1ri;,
L\

voi x,v ciing

dau.

didm;. Tim gi6 tri nhd nhdt cria biiiu thric:

e=-{**o^J*a-Fr[-6"G-9
C6u 3 (,1,0 diiJ@. Tinn tiit cecdc nghigm nguy6n

(*;y)

cira phuong trinh:

Zxy'+x+Y =$J

C6u 4 (; ,A ciiA@. Tim tht cac6c s5 c6 5 chfr sd

otni,

sao cho

|iffi

= ah .

Cflu 5 (1,0 di6m). Cho ba s6 a, b, c nguy€n dlrong, nguy€n t5 cung nhau vd thda
mdn diAu kiQn:

1o 1-

abc

l.

Ch,irg minh a + bli sd chinh phuong.

Cfiu 6 (.1,5 di€m). Cho duhng trdn t6m O vi ddy cung AB.Tii mQt di6m

dunng tritn (M kh6c

M b*ki

tr6n

A vit B), ke MH -t-AB tai 1{. Ggi E, F lAn luqt ld hinh chi6u

vu6ng g6c cua 11 tr6n MA, MB. Qua &/ kd tluong th6ng vu6ng g6c v6i EF, cht ddy
cung AB l:aiD. Chtmg minh

AH
.AD
ring:,. W *

MB2

BD- BH

Cffu 7 ()1,0 di€m). Cho tam gi6c ABC vudng tqi A, duong caa AH. Gpi (O) ld ttudng
trdn ngo4i ti6p tam gi6c AHC. 'llr6n cung rlhb AH cta (O) 16y di6m Mbdt ki kh6cl
vd H.

Tr,Otr ti;5p

tuy6n @i M cua (O) lEy hai diOm D, E sao cho BD

- BE =Bl . Eudng

thing BA4'c}it(O) t4i diem thf hai 1{"
a)

Chrq; minh r6ng tri

gi6c BDNE nQi ti6p dirgc trong mQt
b) Chrmg minh Cucrng trdn ngcpi tiep

tt gi6c fiDlttlvirduong
---- H6t----

tr0n (C) ti6p xirc n-hau.


IITTONG DAN CUAVT

uBND TiNH ruAlNcwBN
sO crAo DUC vAoAo rAo

THI TUYEN STNTT LOP 10 THPT CUWEN
NAM HQC 2oL6 -2017

vtON THr: ToAN-HQC
@inh cho thi sinh thi viro chuy6n Todn)
( Bdn hudng ddn chtim gdm c6 04

trang)

I. Hufng din- chung
- .r . ,. -7 t, r gi6 etung bdi
- Gi6m f.frao rA" n?f"rrng y6u c6u cria huong ddn ch6m dC $6oh,
t6i tta'
lim cria thi sinh. Thf sinh lem c6ch kttat Aep rin n6u U*g "T:h: cti6m

ho?t voi tinh
tlQng, linh
- Khi van dud diry invi thang diem, gi6m khio cdn chri
tfran tran trgng bii ldm cria hgc
.!
: N.6; #rie"rrri tii5t h6a eti6m c6c can phii rlim b6o k*rong sai lQch voi t6ng
di6m vi du-o. c th6ng nh6ttrong toan hQi tl6ng chdm thi'
16 di5n 0,25
- Ei6m toan f,ai ln t6ng Ei6- cria c6c rau rroi trong d0 thi, chAm di6m
vi kh6ng lim trdn.

sinh'

**'

j !

II. Dfp 6n vi than tli6m

Cffu
115

1

Noi d
thi x-0=(O;O) ldnghiQmcuahQ'
0 thi dAt * = tV(k > 0) ta dugc hQ phucmg trinh

;mu /-0
+ NrSu y

*

)3k4 -17k2 +20_0
Irr&'r'-v')=3w -z(tc'-t)
' tc\tc'z+t)_3k
10
l,r(o'r'* y')=loy

* k--t\v k=tf
Oit-,0ndn

)k=Zvk=tE

=2+ x:Zy thay vio (1) ta duo. c !2 =l=
+ (z;t) v (-z;-t) li nghiQm cria hQ'

+

Voi

k

+

v6i

/, =

g =t[i, *,

)
f

hQ phucrng

Diou kiQn:

(

trinh c6 4 nghiQm: "

x29

e
u.F
t, =

Tac6"=@.

tl

vno (1) ta dusc

li
V$y

y=

x>-9

"

nghiQm ctra hQ.

0r25

0.25


D"

"

tfnh;[fi|]

1r0

,rl*F-.Fl

= l,+.

.,:,,

=

.

oau

"-"

,-s13
lr* J

(:+rFXr-.F)

khi AB > 0' n6n

xhy ra

-.Fl =

6

oee-(r-q)

>,0

<+x<

Vfy minP = 6 khi 9,V+x+y --53 a'4xy +?*+2y =t6 e(Zx+l)(zv+l)=rct
Do 767 h s6 nguydn ti5 n€n 2x+l chi c6 tfre n mQt trong

4 sti

D6u

CAU 3

,lt*rl

* lrl

xby ruUri

>

18

!1"+167

Vdi ?*+1-t ta dugc x=0,! = 83.
+ Vdi ?s+!=-l ta dugc x = -1,! = -84.
+V6i ?*+l-767 tadugc x=83,Y=4.
+Vdi 2t+L--!67 tadugc x=-84,Y --1.
Vfly PT dE cho c6 4 nghiQm nguy6n (x;y)li

0.25

+

(o;

Cflu 4

:

s:), (-t; -84), (83; o), (-t+;

-t)

Tac6: abcde-l000ab+

Ta lai
1'0

0.25

c6

l00Om *

Vi

L}0}ab+rar=(*f . EAt m=ab,n=cde , ta

n= m'

)

m'

>tlo}m+

m' >1000

n < 1000 nan m3
m>

32

0.25

(1)

* m(*'- 1000) < 1000

z>33+ *(*'-1000)>33.89
Do d6 m <33 (2)

N5u

*

dugc

-2937 >1000

0.25

(voly)

0.25

Tt

(1) va (2) suy ra m =32
Ydv abcde --323 =32768

Cffu 5
1r0

0.25

a+b
1 1 I suy ra T-:=c(a+b)=qb=ab-ac-b=0
TU:+;=i,
-'a'b c'-'
ab c
(2)
CQng hai vi5 cua (1) v6i c2 ta dusc (a- c)(b - r)=
"'
1

(1)

a.25

udc nguy6n ducrng oira a- c vir b - c thl c'i d'
0.2s
Do d6 ci d, m{ a'- c vit b - c cingchia hi5t cho d ndn d li u6c cria
a, b, c.Theo gi6 ttrii5t a, b, c nguy6n td cung nhau n€n d
U
vit b-c kh6ng c6 u6c lon hcrn 1, n6n (a
Yfly
li s6 chinh phucrng kh6c 0.25
s6 chinh phucrng, tumg sti a - c vd b

Ni5u d

li

-l
-')(b-c)

a-c

-c

nhau.

- c = m' + c' - (o - t)(b - c) = k''m' ) c = hn
> a + b - (o- ") + (U - r) + 2c -- k' + m' + Ztctn = (tc + m)'
YAy a+b ldsiS chinh Phuong.
Ddt a - c = k2,b

2


Ciu

6

1r5

0.25

Gqi E',

Ia co:

F'

lAn luqt ld hinh chiiiu vu6ng g6c ctra D tr6n 1/L4, MB.

M _S*, _ HE.MA , D _S*o _DE'..MA
BD S*o - DF',.MB' BH S*, HF.MB
AD AH T,IA2 HE.DE' /.\
lrr'
BD BH--.--MB' DF'.HF \ /

=frF

Ta c6: ffiF
Ta lpi c6: ffiE
114d,

(

=ffi'

ffi,=ffi,(
..+

vi cung bir v6i

g6c

0.25

0.2s

fuB)

( hai g6c c6 canh tucrng img vu6ng g6c)

cung ch5n cung

DF)

0.25

^
= HFE = DE'F'
vd DF'E'd6ng dang
Suy ra hai ta:n giac
IfD f\DI
HE DF'
ij-I'?:
(2\
+
+ nE.DE' = HF.DF'
=

nrr

=-DE'
HF
-

Tt

(1) vd (2) ta dusc

MAz
MB2

-=

CiuT

HI,"DI''
AH AD
BD BH

0.25

=l

0.25

2r0

0.25

N'

a)

Vi

AH L BC

*fu

=900

+ AC lilduong kinh cria tlucmg trdn

(b). n{a MBC vu6ng t$ A nln AB ln ti6p tuyiSn ctra (O).
X6t c6c tarr_ gi6c BAMvir BNA c6: ffi chung vd ffi =ffi vl
c6 si5 do bdng nria sii do cung fu .YAy MAM vd tau,l d6ng

.BMBA
+;;=#=

dang

0.25

BM.BN = BAz .

Theo gin thiiSt BA= BD + BM.BN = BDz

- *BD= *BN

.1'trl.]:i,i!s+ii;i'i'Ei{i{.'a!qi:{]1i':ri.::i.]:ii..gi!!ia;;}nii}.:ji:'1i-.;,&si,},u$g#d&;.:...,--...,":,;e.,:.

0.25

ffi

Mpt kh6c
ng6c chung crta LBDM vd IAWO .
Vfy n6n LBDM vd tauo ddng dang
Ta c6 BD=BE= LBDE cdntpi B
+ Tri gi6c BDNE nQi ti6p dugc trong mQt cluong trdn.,
b) Ki hiQu ctucrng trdn ngopi ti6p fli gi6c BDNE ld (O)
Gii srl (O) vd (O ) kh6ng tiisp xirc nhau. Vi (O) vd (O ) c6 di6m
chung l/n6n chring c6 cti€m chung thri hai la N' vd BN' cht On tai
di6m M'h,hdc M.
Cdc tam gi6c BDM' vd BN'D c6:
chung, ffiD =frD (hai
,
..1
^.
g6c nQi ti6p cung ch5n cung BD cl&ra (O )),

=ffi =ffi
+ffi =fri+ffi=ffi

ffi'

frD=fiE=ffiD=frE

Vfy n6n cdc tam gi6c BDM' vd BN'D ddng
BM'
BD

uu
""= BD= BN' = BM'.BN'= BDz

dang

Ta c6 : BM.BN = BD2 n€n BM' .BN' = BM.BN

= BN'-

Hai tam gi6c BMM'vdBN'Nc6

ffi'

ffi'= ffi

=+

chring d6ng dang

=

BM

chung9 ud

ffi

0.2s

+

ffi

BM'
BN

n€n
%=Yr'
BN' BN
= t8oo

Tir gi6c M'MNN'nQi tii5p iludrng trdn
0.25
Dulng trdn ngopi ti€:p M'MNN' vd (O) c6 chung nhau ba di6m M,
if, if l n6n chring br)ng nhau ) M' h di6m chung thri hai ct.r- (O)
vd titip tuytin DE. Ei6u niry v6 ly
DiAu gi6 su ld sai.
=
Vpy ducrng trdn ngopi tiilp tri giSc BDNEvi tludrng trdn (O) ti0p I o'zs
=+

I

xtic nhau.

--- ttift----

4

pr cgiNn rntlc

Thli gian lim bii:

Cflu 1 (r,0 dia@. Cho hdm sti
birin hay nghich bi6n tr6n

R.

ruYtx

srNn Lo? 10 rHPT
xAm Hec 2ot6 - 2ot7
vTON THI: TOAN HQC

THr

UBND T1NH THAI NGUVEN
so ctAo DUC vA DAo rAo

120 phfit (kh6ng

,=(J1-Ji)*+z

c6

kA

thdi gian giao

d6thi (d). Hdm sii dacho ldd6ng

? Giai thich? Tim tga tlQ giao di6m crta

Cfru Z (t ,0 di€m). Khong dung m6y tfnh cam tay, rut gen bi6u thric

cnu 3 (1,0 di€m). cho bitiu thric
Hdy rirt ggn B vd tinh

Ciu

,

@)vdi tryc tung.

*(Z*il2[

J;

-#),voi
[w

r=(#,#)
.

-\E8-8

x>o,x*

'

giltri ctr. Bkhi x = 3 +.6'

C6u 4 (1,0 di€m).Xitcdinh c5c he sO avdb, Uititn9 phuong trinh
c6 nghiQm

di)

f**by-4
L[.0, - 2y --2

(*;y) = (z;-l).

5 (1,0 di€m). Kh6ng dung m6y tfnh c6m tay, h5y gi6i phucrng trinh:
xz +6x-2016:0

Cffu 6 (t ,0 di€m). Cho phucrng trinh

x' -

Zmx

* (*'- +):O 1t;,

rn

li

tham

si5.

a) Chung minh phucrng trinh ( 1) lu6n c6 2 nghiQm ph6n biQt vdi mqi gi6 t4 cliua m.
b) Goi x, xrlir2nghiQmcuaphuongtrinh(1). Tim mdd *r'*xz2 =26'
Ciu 7 (1,0 di6m). Kh6ng tinh ttmg gi6 tri cp th6, hiy
cos

20' , sin 3 8' , cos 5 5' , tan

480

sdp x6p c6c

ti s5 lugng

gi6c

, sin 88' theo thri tg tSng dAn, gi6i thich?

Cffug (l,0diAm). Chotamgi6c ABC vu6ng tU A,c6 sin.B=l.16rtinhc6ctisiS luqng
gi6c cira g6c C.

Cfiu 9 (1,0 di€m). Cho ttuong trdn tdm
tirip tuytin AB,
song

trdn

AC voi

Ciu l0

di6m Anim ngodi duong trdn. K62

duong trdn (^B,C ld2ti€p diCm). Qua C k6 mQt duong thEng song

voi OB, cgt OA taiH. Chrmg minh
vi I/

O vdmQt

rEng

tu

giac

ABOC

nQi tiep duo. c trong mQt duong

ld @c.t6m cfia tam giac ABC.

(1,0 diA@. Cho

tf

gi6c

ABCD

nQi ti6p tiong

ilulng tron(O;R),c6 hai duong

ch6o vu6ng g6c v6i nhau vd ciltnhau tpi /.
a) Chung

minh.i,g, IA.DC -

ID.AB;

b) Tinh t6ng AB2 + CDz theo R.

--- rudt---S6 bao

danh:

...i!,.


UBND TINH THAI NGUYEN

HIIONG NAN CHAM
THI TUYEN SINH VAO LOP 10 THPT
NAvr Hec zot6- 2ot7
vrON THr: ToAN Hec

so GrAo pvc vA pAo r4.o

( Bdn hudng ddn chtim

gim cd 04 trang)

I.Iluring din chung
- Gi6m kh6o cdn ndm virng y6u cAu cria hudng d6n ch6m Oe Oantr gi6 rhing bdi
ldm cta thf sinh. Thi sinh ldm c6ch khric d6p dn n6u dring v6n cho di6m tOi rla.
- Khi vfln dUng dbp in vi thang di6m, girim kh6o cAn chri tlQng, Iinh hoat vdi tinh
thAn trdnlrgng bdi ldm Lta hgc sinh.
- NiSu c6 viQc chi ti6t h6a di6m cdc,y cAn ph6i ddm b6o kh6ng sai lQch vdi t6ng
di6m vi - Di6m todn bdi ld t6ng dii5m cria c6c c-6u h6i trong dC thi, chSm di6m 16 di:n 0,25
vd kh6ng ldm trdn.
II. Dip rinn vi thang tli6m
Ciu
NQi dung
Di6m

.
.

CAU

I

Cho hdm s6 y=

(J1-Jr).+3

c6 Ao

tni (d). Ham so da cho ld ddng

biiin hay nghich biiSn tr€n R ? Giai thich? Tim tga d6 giao cti6m
c:ioa

(d)voi truc tung

Gidi:

, =(J1 -Ji)*+S

Hdm s6

Vi

I

0,25

ddng bitin tren tR ,

a: Jl-Jlro

Cho x :

0 ta c6 y :3

0r25
suy ra toa dQ giao diiSm cria (d) v6i tryc tung ld 0r5

A(0;3)

Ciu

2

Kh6ng dung miry tinh
thtrc: A

-

+rJr)'-

(z

..f&8

cAm

.

Gidi:
Ta c6:

e-(z+sJz)'-.,D88

-* +2.2.3,12+1l,li1z -,!1aat

0'5

=4+lzJr+18-i2J,
-11
Cffu 3

chobi6u

,no",r-(

x)0,x*4.

G-!.:L ).(J--Gl') ,,-F)''-

0r25
0,25

[;;5*G-/[m

Hay rut ggn B vd tinh gi6

trf

cta B khi x

-

3

+

.6.

Gi,rti:

ruvv'"-[cG5-Gi)t@]
*Y.i-l [v;'tr-('tr- zXv;.2)l

ra c6:' =(

o(G-t) l;$r._z)

-Wq'
_

0r25

0r25

4
1


0r25

vdi r-3+..6=
Ciu

(Jr.,)'= J; -J-z*t:+.8 -J|+t-1 =J,

0,25

4

Xric tlinh c6c hQ s6 a vd b, bilth6 phuong trinh
c6 nghiQm

(*;y)

-(z;-t).

Gidi:

Vi he c6 nghiQm (2,-l) n€n th6 x=2, y:-l
lzo+b1-r1-4 lzo-b-4

{

-2.(-r)

lzb

I
={

Ciu

5

--z

lb

c6:
0,25

--z

a-l

lb

+ Vdi

<>{

vdo he ta

0,25

--2
a:l, b: -2 ta c6 hQ

l*-2y'

0r25

lr.-,
(2;-l)

0,25

Vfly de h0 c6 nghiQm (x,' 1:
Kh6ng dung mdy tfnh cAm tay, hay gi6i pnuong trintL

x2

+6x-2016:0

Gidi:

a:1, b:6 (b':3), c:-2016
L = 32 - 1.(-2016) = 2o2s> 0,l[

Ta c6

= 45

Phuong trinh c6 hai nghidm ld: x, = -3 + 45 = 42;xz = -J * 45

Ciu

6

Cho phucrng trinh

x' -zmx *(*' - +):O 1t;, z

ld tham

- -48

0r25
0r25
0r5

si5.

a)Chung minh phuong trinh (1) lu6n c6 2 nghiQm phdn biQt vdi mgi

gi|tri

c: ia m.

b)Ggi x11x2li 2 nghiQm cria phuong trinh (l). Tim m di: xr2 +t =N,
Gidi:

a) Ta c6: A =m'-(*'-4)=4>0,Ym

suy ra phuong trinh lu6n
c6 hai nghiQm phdn biQt voi mgi gid,tri m.

b)

Theo viet ta

.o' {

xt + x2 =

[','',

"'*
-4

= m'

+4 =(x.,+xr)' -Zx,.xr=12*)' -2(*'-4)=2m2 +g
TheobAi xf +4=26+2mz +8=26*m=*3
Thri lpi: Vdi m=3, ta c6 phuong trinh x'-6x+5:0 c6 hai
nghiQm: x, =1,x, = 5 =+ t * 4 - 26,thoa m6n dAu bdi
ygl_!! - -3 cfing th6a m6n itAu bdi. Ydy m = +3

0,25

0r25

Ma xf

2

0,25

Ar25


Cdu 7

Khdng tinh tirng gi6 tr! cg thil, hay sip x6p cric ti sti lugng giitc
cos20', sin38', cos55', tan480, sin88' theo thti tg ting ddn, giii
thich?

Gidi:
0r25

= cos52o ,sin 88' = cos 20
Vdi 0 Ta c6 bi6n cl6i sau: sin 380

0,25

ndn 0N6u 450 < P <900 thi tanp 21 = tan480 > I
Do d6 ta sEp xiip theo thri tU ting dAn nhu sau:
cos

CAU 8

55',

Chotam

sin 38", cos

0r25

20', sin88o, tan 480

ABC vudng

gi6c

0r25

tqi

A,c6 sin B -:.HEy tfnh citctisri luqng
3

Gidi: (Kh6ng
Ta c6:

C

C.

giric cria goc

cho diiim hinh vd)

llA

sin,B=1=)CosC=-

^^
JJ

B

ACIAC2IACzlACl
- BC 3- BC2 g- AB'- 8 - AB 2J,

I
haycotg
=-|-,suy ra: tanC =ZJi
,
2J2,

0r25
0r25
0r25

AB

AC

Ciu

9

Cho dudmg trdn tdm

duong tlrang song song

ABOC

O vi

mQt

di€:m

AnAm ngoii duong trdn. Kd 2

AC voi ducrng tdn (B,C ld 2 n}p di6m).

tiiip tuytin AB,

0,25

',=+
Qua C kd mQt

vu OB, cfut O,l t1i H. Chung minh rang th gec

nQi ti6p duo. c trong mQt duong

tdn vd f/li trfc t6m cua tam gic

ABC.
Grrii.' Flinh vE

Xdt

tu

trEd

:

gdc ABOC

=frd

=eoo (Theo

tinh ch6t cta ti6p

tuyiSn).

Suy ra

tu gi6c ABOC

ti6p

c tong mQt ducrng

du-o.

0,25

c6:

0,25

nQi

tdn.
a

J

. {lca tloB +CHIAB

Trlc6:

Trong tarn gircABC c6:

Ciu

10

Cho

0r25

loB L AB

{':,:
11=H h tuc tam cua tam gi6c ABC.
IAH L BC

tt gi6cABCD nQi ti6p trong ducrng trdn(O;R),c6

ch6o vu6n

a)
b)

g

g6c

0,25

hai dudmg

vdi nhau vd cit nhau t4i L

Chrmg minh ring: IA.DC

- ID.AB;

Tinh t6ng AB2 + CD2 theo R.

Gidi:

Hinh vE kh6ng cho tli6m

a) Xdt hai tam gi6c IAB vd IDC c6:

vd

fri =i6Z (g6c nOi tii5p cung chiin

cung

0r25
D

BQ

Suy ra: NAB diing d4ng

voi NDC

do

0r25

.IA
rc +IADC=ID.AB
-ID =fi

oo ta co.

b) Ke ducrng ktuh CE cta duong

tdn (O).

ili =ide =90'(g6c nOi ti6p chan duong kfnh EC)
AE t tBD+ ABDE le hinh thang cdn (hinh thang
Suy ra {n:"!*
'
LAE

0,25

=(zn)' =4R,(DoADEC

0r25

Ta c6

IAC
nQi titip duong trdn)+ AB = DE
Do d6 AB2 +CDz = DE2 +CD2 = ECz
vu6ng tai D).
YQy AB2 + CDz

= 4R2

-- ttd+

4


,,

THI TUYTN

UBND TINH THAI NGUYEN

VAO LOP 10 THPT
NIT hsc20l7 - 2018
MON: TOAN HQC

sd cra.o uuc va oAo rAo
on cniNn rHUc

,

SWTT

Thdi gian ldm bdi: I20 philt, kh6ng kA thdi gian giao
(Dd tui gdm c6 01 trang)

CAu 1(1,0 di6m). Kh6ng dung m6y tinh cdm tay hey giiriphuong trinh: xi

Ciu

2(1,0 tli6m). Cho him

siS

Uac rirrdt y

=(2m4)*+5m-l (m

lirtham

+2x-8

di

= 0.

,0,**)1.
,2,

Tim m d6hdm siS nghich bi6n tr€n IR. .
b. Tim m d6OO ttri hnm s5 c6t trgc tung t4i.tliiSm c6 tung tIQ h -6.
Cfiu 3(1,0 tli6m)' Kh6ng dtng m6y tinh cAm tay, rut ggn bi6u thrlc:
a.

, = (S - zJ-z + 2{s)(Jr+ 1o@)

.

+-

L+t *6x+J-xl,f{_:.-r) nai I'=0.
[Jx+3 Jx-3 x-e J(Jx+: ) [x+e

cho B =(

vitinh ei|tricria B khi x --12+6Ji.
(mx-v=n
cau 5(r,0 tli6nr). cho h€ phuong
li tham si5).
HEyrutggnbi6uthric.B

"r* IT;;:r(m;n

a. Khdng

dtng m6y tfnh cim tay hay gi6i hg phuong trinh khi

b. X6c itinh c6c tham sd m

* --!;n
2'3-1.

vi n biltring hQ phuong trinh c6 nghiQm li (-Lr6).

CAu 6(1,0 tli6m). Cho phuong trinh 2x2+3x-l=0. Ggi xr;xz li hai nghiQm phAn biQt cria
phuong trinh. Kh6ng gi6i phuong trinh h6y tinh gi6 tri cria bi6u thric:

!r*L-\.
[r, \)

p =z(

Cflu 7(1,0 ili6m). MQt tam gi6c vu6ng c6 c4nh huyAn bing 5cm, di$n tich

li

6cm'. Tinh

dO ddi

c6c c4nh g6c vu6ng cria tam gi6c vuOng d6.

Cfiu 8(1,0 tli6m). Hai dulng trOn (O) va (O') cit nhau t1i
OO' . Qua ,,4 k6 tluong thang vu6ng g6c
vir

D.

D

vir B. Gqi

M

ldtrung di'5m cta

vdi AM cit c6c dudrng trOn (O) va (O')

tan tugt O C

ring AC = AD
tli6m). Cho dudmgtrdn (O), Auongkinh AB,ctxrg D ni*cirngphfa A6ivA ,lA

Chung minh

Cffu 9(1,0
(

A

thuQc cung nh6
a.

fu ).Gqi E ld giao eliiim ciua AC vir BD , F ldgiao diiSm ctn AD vd BC .

Tinh g6c frE khi

siS

r*g D

do cria

"*g

ffi

D

bing 800.

.

khi g6c
ailne ss' ,
CAu 10(1,0 tli6m). Cho tam gi6c nhgn ABC (ABcqrrh AC, AB hn luqt t4i D vir E. H ld giao diiSm cria BD vit CE, K li giao itii5m ctn DE vd AH, F lit
giao cti€m cliua AH vir BC . M tdtrung tli€m oiua AH . Chtmg minh rang, MDz = MK -lfiF .
b. Tinh

st5

do


of rru rutl ruytN

so crAo nuc vA DAo rAo rsAl ncuynm
TRI.TC}NG THPT GHUYEN

SINH

vao r,Op to

NArr Hec 2ot6 -zot7
MOn: TOAN TIN

(Ddnh cho th{ sinlt thi vdo chuyAn Tin)
Thdi giun tdm bdi: 150 phtit, khilng kd thdi gian phtit

Biri I (2,0 tli6m). cho

bir6u

&*$).H

thric n=(
(V2" +b +^,t2a-b

^14a,

-b, -2a+b )

^t4a,

di

-b, "u,

2a> b>0.
a. Rrit gen A.

b. Bi6t 2a-b=1. Tim gi6(_nh6 nh6t cria bi6u th{rc A.

Bni 2 (1,0 di6m). Cho phucrng trinh (x+2)(x+3)(x+4Xx+s) =a. Bi6t phuong trinh dd cho
c6 4 nghiQm ph6n biQt x,,x,x3,x4. Gqi

r : xt.x2.x3.x4. H6y tinh P theo a?

Bni 3 (2,0 tli6m). Chrmg minh ring tu zAfi sii nguyEn duong U6t tci lu6n c6 th6 chgn ra
dugc hai s6 md t6ng hofic hiQu cria chfng chia h6t cho 4030.

Bni 4 (2,0 di6m). Trong mat phing cho 2016 itulng trdn trong d6 b6t kj,hai dudmg trdn
ndo cfing

cit

nhau t4i hai cti6m phdn bi6t vd kh6ng c6 ba tlucrng ndo c6 ili6m

chung. Tinh s6 phAn mAt phing md cilc tlulng trdn d6 chia ra. (C6c phAn mat
pheng n€u 6 dAy khdng chdng 16n nhau.
cludmg trdn

thi

Vi dg: n6u tr6n mAt phing c6 dring I

tlucrng trdn d6 chia mflt phing thdnh hai phAn; n6u tr6n mflt phing

c6 dring2 dutrng trdn

clt

nhau t4i hai tli6m phdn biet thi 2 tluong trdn d6 chia mflt

phing thanh a phAn...).

Bii

5 (3,0 tli6m). Cho n ld mQt sd nguy€n duong

chin lcrn hcrn hoflc bing 4.Tat6 mdu

mdi sti trong c6c sti nguy€n duong tu I d6n

mdu

'

xanh

ui, + sO con lqi dugc t6 miu
2

,,l

sao cho

i

sti trong chirng clugc t6'

d6. Vdi m5i c6ch td nhu vfly, kf hiQu

le sii cdc s5 nguy6n duong b5t ki mi n6 c6 thti vitit dugc du6i d4ng t6ng hai s6
khSc mdu.
a.

Tim tdtcitcdc gi6t4ci,r- fo.

b. Chrmg minh rdng

l,
Thi sinh khdng clugc sfr dgng
Hg

vi

tOn

thi

sinh:.........

t

tii liQu. Cfn bQ coi thi kh6ng girii thfch gi th6m.
; Phong thi:......; Sti Uao danh:........

f


oAp AN ur rrU TUUTUYEN SINH

so GrAo DUC VA DAO TAO THAr NG

vAo r,6r

TRT/ONG THPT CHUYEN

ro NAna

Hec

zot6 - zotT

M6n: IOAN rmr
(Ddnh cho thi sinh thi vdo chuyAn Tin)

oi6m

DAP AN

Bei

Bii I

(2,0 rli6m).

2a>b>0.
a)

R{t

ggn A.

b) Bi6t Za-b=1. Tim gi6

tdah6 ntit cta U6u ttrnce.

.

Gidi:

+) Voi 2a>b >0, thi ta c6:

1'0'
1.a

2^l2a+b

t- Y-E "'2a+b-(2a-b) ,{W
u.-

-N-q

4a2 +3bz

4a2 +3bz

=+) Ta co 2a: b

,
A=

1.b

b

* /. Suy ra:

(b+I)? +3b2
=4b+!+2.
--u'
b
b'o

Ap duos b6t I

+u+!r

si5

duong ta c6:

r^E =0.

b\b

Y$y 4>-6. D6u beng xiry rarr,i

lla

=b+t |

]+a

=| * I . i

=

3

lzoruro lo= z
ltla=vay gi6 tri nh6 nhdt

cria

A te 6 dat dusc un,
1u =

t2

1.

1r0

'

Bni2 (1,0 di6m).
Cho phuong
nghiQm

trinh (x+2)(x+3Xx+4Xx+ 5):a. Bi6t phuong trinh tli cho c6 4

phin biQt x,,x,

x3,x4.

Ggi

P

=\.x2.\.x+. Hiy tinh P theo a?

Gidi:
+) Ta c6:

(x+2)(x+3Xx+4Xx+ 5)=

e

(*'

o

0)

+7 x +10).(xz +7 x +12) = a

+) Df,t t = x2 +7x+l 1. Ta dugc phuong trinh An t:

(r-lXr+l)
Theo

2

-aetz -l=a.

gii thi6t, phuong hinh (1) c6 b6n

nghiQm ph6n biQt n6n phucrng trinh (2)

It,+tr=g
It't' = -l-

)1
l.

+) Kh6ng rndt t6ng qu6t, gie sri
x2

+7x+ll-f,

= 0;

\,xo

0r5

(2)

x12x2

a

li hai nghiQm cria phuong trinh

ld hai nghigrn cria phuong trinh x2

+7x+11-tr=9.

Khi d6,

=ll-t.'
{'
\r
t z' {
L4.ro=ll-t,=x,.xrxr;Eo=l2l-ll(tr+tr)+trtz=120-a.
I

x,.x^

015

Vfy ta c6:

P

=120-a.

Bni 3 (2,0 tli6m).
Chfi'ng minh

ring tir

2017 s5 nguyGn duong UAt t
hai s5 mi t6ng hoic hiQu cria chrfing chia h6t cho 4030.
Gidi:
'tchi rhiu mQt s6 nguy6n duong U6t
3

ti

cho 4030 thi c6c sti du

phii

thuQc t?p {0,

t,...,4029).

0r5

Trong 2017 sd tr6n ta chia vio tring nh6rn sau;
+ Nh6m

thf 1 g6m nhtng sti thi chia cho 4030 c6 sti du le 0.

+ Nh6m thL }gdm nhting sti ttri chia cho 4030 c6 sti du

li

I ho{c 4029.

+ Nh6m tht 2015 gdm nhtng sii ttri chia cho 4030 c6 s6 Au U 2014 ho{c 2016.

1r0

"'i


+ Nh6m

Tdt

tht

c6

cir

2016 g6m nhftng s6 khi chia cho 4030 c6 s6 du le 2015.

2}fi

nh6m. V4y trong 2017 s6

nh6m n6n th6a m6n y6u ciu

bii

di

cho c6 2 s6 cirng thuqc mat

to6n.

015

Bii 4 (2,0 tli6m). Trong mit phing cho 2016 tlulng trdn trong d6 bAt kf hai
dulng trdn niro ciing cit nhau t4i hai di6m phin biQt vir kh6ng c6 ba du'&ng
c6 tli6m cfiung. Tinh s6 phin

nio

(C6c phAn

c6 dring

mit phing mi cic tlrd'ng trdn d6 chia ra.
m{t phing n6u 0 ddy khdng ch6ng lOn nhau. Vi dU: n6u tr6n m[t phing

1 ducrng tr6n

thi dulng tron d6 chia mflt phing thdnh hai phin; n6u tr6n

rn[t phing c6 dtng 2 ducrng trdn cit nhau t4i hai ditim phin

biQt

thi 2 dudng tron

d6 chia m[t phing thinh a phAn...).

Gidi:

bii

+) Ta xdt

biit

tt6

kj,

torin t6ng

hai fuong

qrofut:

tdn

"Trong mqt phdng cho n (nel$'

) firdng trdn trong

ndo ciing cdt nhau tqi hai di*m phdn biet vd kh6ng cd ba

dudng ndo c6 diim chung. Tinh sii phin mfit phing md cdc iludng trdn d6 chia

"

ra.

Y

4

+) Gqi Pt

li s6 phin mit phdng dugc chia ra b&i ft dulng trdn trong d6 b6t k! hai

tlunng hdn nio ctng cit nhau t+i hai rtiiSm phAn biQt
di6m chung.DA thily

khdng c6 ba t1u,ilng nio c6

P1:2; P2:4.

+) X6t tludrng trdn thri (k + l):theo
I

vi

gii

thi6t, & tlucmg hdn cdn lqi ciltdulng tron

tht

(k + 1) tqi 2k giao tti€m. C6c giao di6m niy chia dudng trdn thr? (k + 1) thenh

2k

cwrg. M5i cung

niy

chia ph6n mflt phing n6 tli qua thinh hai phen. Do tt6 s6

phdn mflt ph[ng moi sinh ra do c6 th6m duong trdn thri (k +

Vfy

ta

c6:

SuYra ,

Po*r'=.Q

+2k,Vt

eN'.

Pro,u

I) h 2k phin-

i

4=nz -n+Z,Yn €N'.

v0Y

1r0

=2a162 -2016+2=4062242.

1r0


./,
'r4
Bni 5 (3,0 tli6m)

Cho n lir mQt

s6 nguy6n

dutrng chin kfin hon ho{c bnng 4.TatO rniru m5i

tt 1 tl6n n sao cho I2

trong cic s6 nguyOn duong
xanh

s5

trong chrfing tluqc

tO

s6

miru

vi ;,1.
s0 cdn l?i tluqc tO mdru d6. VOi m6i cich tO nhu v4y, ki hiQu .f In
2
a

, r.

sd^ cic sO nguy6n

duong -r.-.
UAt t
dufi

dgng t6ng hai s5

kh6c miru

Tim tdtcitcitcgiirtricrtr fu.
b. Chrrimg minh ring loo <1g7.
a.

Gidi:

I tlugc t6 xanh. Ta c6 c6c trulng

+) Kh6ng rn6t t6ng qu6t, coi
5a

IXZI*

(suy ra 3D 4D)

: fo=3 (cic

si5

hgrp sau:

vii5t dudi d4ng t6ng cria hai s6 kh6c mdu ld
1r0

4;5;6);

lX 2D 3X (suy ra 4D) :' f* =3 (cic sri vitit duoi dang t6ng cira hai s6 khdc mdu li

3;5;7);

lX 2E 3D (suy ra 4X)

.fo =

4

(c6c s6 vi6t duoi d4ng t6ng ctra hai s6 kh6c mdu ld

3;4;6;7);

1r0

Vav

+ ) Rd ring

f, e {:;a}

l*

<197,do c6c s6 c6 th6 vitlt duoi d4ng t6ng cria hai s6 kh6c miu

lu6n l6n hcrn hoflc beng 3 vd nh6 hon hoic bing 199.

, Ntiu f,*=l97thi tu 3 di5n 99 i16u vi6t
5b

rniu. Vd do d6, ta c6 th6 coi

dpng t6ng cira hai sti kh6c

l)i

ducr.c

dudi d4ng t6ng ctra hai sti kh6c
1r0

2D.L4ivi c6c sii tu + diSn 99 cfing vi6t dugc dudi

miu n6n 3D,4D,...,98D. Lric niy, c6c s(i tlugc t6 E

vust qu6 50 s5 (v6 lf). Vfly ta c6 tliAu phii chring

Gi6m khio chAm biri chrft f: Ndu thl sinh gidi bdng cdch khric md drtng thivfrn cho itiim tdi ita

_sr


uauvEN
r4o

nUoxc oAN cnAnn

uBND riNH ruAr
sO crAo nuc vA BAo

rnr ruytN

sINH vAo LOp 10 THpr
NAnn Hec 2ot7-2olB
vt0l THI: ToAN HQC

ddn chdm

r. Hu6nrg

gim c6 06 trang)

ua^?iar!"*r

- Gi6m kh6o cdn nim vrng y€u cAu cria huong O5n ch6m
ldm cria thi sinh. Thi sinh

lim

c6ch kh6c

e16p

O6 Aann

gi6 etung bni

6n niiu ttung v6n cho diiSm ti5i ea.

- Khi vfn dpng d6p anvi thang ditim, gi6m kh6o c6n chri dQng, linh ho4t voi tinh
thdn trdn trQng bai lam cria hgc sinh.

- N6u c6 vi€c chi tirit h6a eli6m
di6m

vi

cdc

y

cdn ph6i e16m b6o kh6ng sai lQch vdi t6ng

vi dugc thting nh6t trong toan hQi tl6ng ctrrim Ai.
- Di6m toan bii ld t6ng eti6m cta c6c cdu h6i trong d6 thi, ch6m di6m 16 eliSn 0,25

kh6ng lim trdn.

II. Din 6n vir thans
a
tli6m
Cffu

I

1,0

tIi6m

Ciu

2

1'0
tIi6m

Ta c6

A'=

- 1.(-8)

= 9 suy ra J L, -3
Do d6 phuong trinh c6 hai nghigm phdn bi-6t

0r5

-l-3
,,= -1+3=.l,va*r=T=-*
1

0r5

a.

Him

sO

2m-3<

12

nghich bi6n tr6n

o <+

R khi vi chi khi
0r5

*.12

b. DO thi him sO cat tryc tung tai

e1i6m

Sm-l=-6e m=-l
CAU 3

Di6m

NOi duns

CAU

c6 tung dO h -6 khi vd chi khi
0r5

Ta c6:

, = (Jt - tJ-z * zJi)(Jr+ 1o,@)
1r0

-.t
(lrem

=(rJi 4Jz +2Ji)(Jr.6)
=(rJi

-l;)(zl;

*lr)

0125

0r25

=(rs)'-(Jr)'

0r25

=18

0,25


Cf,u 4

Ta c6:

a =(

+-i=tf-

ax+

Jil,[€-' -,]

1'0

tli6m

,(G-E)-(,+r)("6+:)+0,+G .Ji -t-Ji -z
Ji+t
xJi -3x- G -3 +6x*G G+r
*Ji
='4 -3x _
_

0,25

0r25

Ji +t
-3
=(J;;)-(GL)
4
I
=zlJ x

-t)

0r25

Khi x =t2+eJ1=(t*Ji)' tac6 G =3+Jl
r
1J3
Khid6 B=
6
z(t +J, - r)

zJt

Ciu

5
1,0

-.t
olem

a.

Khi

* - -!rn= |

2'3

0,25

,u c6 hQ phuong trinh de cho tr& thanh

[r
l-r.-Y-i e(
<
^l-t*-6y=z
-12'-3v-6
Itr-1r=r
Lr 2'
1

0r25

l-t*-6y =z -f'*=:o - ^1.=+
<+i_
Zx-6e<
- =_*
- lr=?
- 14, -6y=12-^<+1
|.,

0,25

b. Vi hQ phuong trinh c6 nghiQm'(-U.,6) n€n ta c6:

l-*-Ji=n ^[**n=-^.6^ I(",6 +t)m=t-.6
- l,lz*-n-, - [, --nL-Ji
L-r* Ji*-r

*{*=# *l*=
ln

-

-m-J3

l,

=

('i{)'
-*-Ji

*{*=

! -r,

0r25

0,25