Đề Toán tuyển sinh vào lớp 10 THPT năm học 2019 – 2020 sở GDĐT Hà Nội | Toán học, Đề thi vào lớp 10 - Ôn Luyện

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ri

rru ruyBx

srxn r,op

to

rgpr

NAU

ugc

2ot9 _-2020

M6n thi : TOAN

Ngdy thi : 02 thdng 6 ndm 2019

Thdi gian ldm

bii

: 120 philt

Biit

(2

Cho hai

4

J".r)

25-x

vitB=

15-J;

-+

r

_-25

t

-T=

_

vu+t)

Vz + 1

-T=-!u -

b

vo,r>0,r+25.

l) Tinh gi6 tri cria bi6u thric ,4 _{khi o =}9.

2) Rrit gqn bi6u thric B.

3) Tim t6t

ci

gi6 tri nguyCn cria o d6 biAu thrlc P ₌A.B dgt gi6 tri nguyOn lon nhdt.

BnifiQ,5

diiim)

1) Gidi bai bdn sau bing cdch lQp phtrong

tinh

ho\c hQ phtrong trinh :

, Hai dOi cdng nh6n cirng ldm chung mQt c6ng viQc thi sau 15 ngdy

lim

xong. Ni5u ttQi thri
nhAt lim ri6ng trong 3 ngay rdi drmg lai

vi

<IQi

tht

hai ldm ti6p c6ng viQc 116 trong 5 _ngdythi

ci

hai tl6i hoan thanh itugc 25vo c6ng viQc. Hdi niiu m5i dQi ldm ri€ng thi trong bao nhi€u ngdy moi

xong c6ng vi€c trdn ?

2) MQt b6n nu6c inox c6 d4ng mQt hinh trg voi chiAu cao L,75

m

vitdiQn tich dity ld 0,32m2.

Hdi b6n nu6c ndy dyng dAy duqc bao nhi6u mdt kh6i nu6c ? (86 qua b,! ddy cua b6n nudc).

Birillr

(2,0 didm)

1) GiAi phuong trinh c{

_-

7 a2

-

78 ₌0.

2) Trong mIt phing tga dQ Oq,cho duong thdng (d):y ₌27.,o

-

*'? +

I

vd parabol (P):y _{= ,z}

a) Chrmg minh (d) lu6n

cit

(P)

_ai

hai di6m phan biQt.

b) Tim tdt ce gi6 tri cria rn itil (d)

cit

(p)

tai hai di6m ph6n biQt c6 hodnh dg

q,r,

th6a

_1t

-2

man-+-=-+1.

0, tr,

arr,

BniIV

_Q,0di1m)

Cho tam giitc

ABC

c6 ba g6c nhqn (,48

<,4C)

nQi ti6p duong trdn (O). Hai dudng cao

BE

vd CF cua tam gi6c

ABC

c*tnhau tai diem 11.

1) Chwrg minh b6n di5m

B,C,E,F

ctng thu6c mQt tludng trdn.
2) Chrmg minh <ludmg thing OA w6ng g6c voi <ludng thing EF.

3) Gpi

K

ld trung <li0m cira do4n

thing BC.

Dudng thing

AO cit

duong thing

BC

tai
di6m.I, tluongthing

EF

citduong

thing

AH

tai di6m P. Chrmgminhtam giitc

ApE

ddng
d4ng voi tam gidc

AIB

vit<luong thlng /(11 song song voi duong thing .IP.

Bniv

@,5 diim)

Cho bii3u thric P = aa +ba

_-ab,

v6i a,b lit cAc s6 thgc th6a

min

a2

+b'

+ab

_=5.

_Timgie tri

lcvn nh6t vd gi6 tri nhd nhdt cria bi6u thric P.

Cdn bQ coi thi kh6ng gidi thich gi tham.

Ho t6n thi sinh 56 b6o danh

Ho tdn, chir ki cua crin bQ coi thi s6 1 :

HCt

H9 t€n, cht ki cria c6n bQ coi thi s6 2 :

so

GrAo DUC

vA DAo

rAo

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so

GIAO DUC

vA EAo

rAo

HAN

Ki

Tr{r

ruyEN

srNH

vAo

Ldp

t0

THpr

NAM 2019

oAp AN

_-

THANc

DIEM
Mdn

thi:

TOAN

rDap

itt

Thungt Jietn gom [)-! rrattgt

+(.6.

r)

I

Irii

\'

Tinh gi6

tri

cta bi6u thri'c A =

khi:r=9

/) - _t:

1

6*r

Thay z

:

9 vdo bi6u thuc ,4 =

25-9

Tinh duoc ,4 = 1

Rrit gqn bi6u thri'c B

-

I f-

va

r>o.t

*2b.

I

Vz-5

15-

I

t25

2

Vr+5

15-

_"'6

2

tl-G*z(

_"t

i)

+

t:

_-25

1

J"-s

.6*s

(.'t.s)(.,6-s)

BNi

I

2,0 diam

l)

r

1

Vr-5

t_t-V.r-5 V.r+I

Di6m

l)

n

)\

0,25
0,5

0.25

n

)\

n

)\

n 15

1,0

2)

D1

Vr+1

T

nguy

im tat c1i girl

tri

nguyOn cfra .r d

6n k?n nhAt.

bi u thric _{P = ,4.8}dat gi6

tri

0,5

3)

+(G.r)

_1.4

".f,*r

_25-r

P=A,B=

n

)\

25-t'

I-ap lu{n d0 .p dat giii n'i nguy6n lon nhAt bang .1 _khi_:r: _2.1

llai

ddi c6ng

nhin

ctng lhm chung rnQt cdng vifc thi sau

1l

nghl'lirm

xong. N5u dQi thri'nh6t tirm ri6ng trong

;l

ngav r6i dnng 14i vir d6i thrir hai

lirm ti6p c6ng viQc d6 trong

5

nghy thi cri hai tlQi hohn thAnh dw1c 25%:

cdng viQc. Htii n6u m6i a6i lhm ri6ng thi trong bao nhi6u nghy m6i rong

c6ng viQc tr6n ?

cqi ihoi gian doi tnu nhdt ti,r, rieng xong.ong

"ie.

li

;

ia*

ii'

;8ir.

.;

;

aii

Goi thoi gian ddi thu hai ldm ri6ng xong c6ng vi6c ld y (d<m vi: ngdv, u > 0)

I

ngAy dQi rhu nhdt l2rm duoc

]

tcong uiect

:t

I ngal dQi thu hai lirm dutvc

I

(c6ng viec)

v

0,2 5

2,0

o:25

o )5

Bii

II

2.5 cli€nt 1)

n )1

Hai d6i ctng

lim

chung mQt c6ng vi6c sau 15 ngdy xong. c6 phuong trinh

L

NQi dung

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1tt

-+ =

_{r 1!}

₁₅

(l)

n

)\

0,2 5
DQi thu t

lim

ri6ng trong 3 ngay

vi

d6i thu hai ldm tlep cong vrec d6 trong

5 ngiy thi ca hai dQi hodn thdnh duoc 257,, c6ng vi6c, c6 phuong trinh

!*!=11:i

rtl

4

Tu ( I ₎va (2) ta cti h6 phuong trinh

111

:r ll

15

351

:x?)1

Gidi hC phuong trinh tim duoc

!=40

D i chi6u clidu ki6n vd k€t luAn: d6i thir nhdt Iim ri6ng thi sau 24 ngdy xong

n

)i

n

);

o

n )5

con vi c. cl i thu hai lim ri6 thi sau .10 n a xon c oll VICC.

2)

l)

#

:dG

via

\

BNi

III

2.O dretn

2)

b) Tim tat ce

gi{

tri cta

m dc (d)

cIt

(P) tai

hai.li

m

phin bift

c6

11

2

n+=+

TT

c

hoinh

aI6 :r

.:r

thiia

mi

'll _r 1

rr"

Theo dinh ly Vi-et:

rt

+ 12 _{= 2rn}vd

r,r,

_{= m}
tla

'+I:DK: .r{ /Oatnt'l

J' J, r,J',

-1

tich ilrly

li

0.32 nr'.

tkii

bdn nu'6c

niv

tlung tl6y du'gc bao nhi6u m6t kh6i

nu'oc ? (Bri qua bO

diy

ciia b6n nu'6c).

56 m6t kh6i nu6c bdn dirng ddy bAng th6 tich crha bdn.

Bdn nu6c drmg duoc s5 m6t kh6i nuoc ld 1. 75 x 0. 32 ₌0. 56

(m')

0,25

n

);

n

)i

'l'H1:

_r"

₌

_2(v6_nghiQm)

TH2:

r'

-

9e

:" _{= 3}hoac

r=-3

n

)\

9
) 0

0,5

(z'

+

z)(

NI6t b n nu'6c inox c6 d4ng m6t hinh trg v6i chi u cao 1.75 rrr vh diQn

0.2 s

t,0

T4p nghiQm phuongtrinh S = J

I

l

't.rong

_m{t

_phing_toa_d0_Or;u._{cho tlu<rng}

_thing

_(d):'u₌_2nt,.r

_-

_nt,'₊

_I

vlr parabol

(P):u

-

.r-'.

a) Chring minh (d) lu6n

cit

(P) tgi hai tli6m

phin

bi6t.

Xdt phuong trinh hodnh d6 giao di6m cta (d) vd (P)

:t'

=2ntr

-m,'

+L

e

tt -2m^r+m'-l=O

L.=m

_{-(. -r)=1:A}

_\./

>0

Vm

Vay (,1) 1u6n

cit

(P) tai hai di0m phan birlt vrii moi gi6 tri cua rn.

015

0,2 5

0,2 5

0,s

0,5

Gi6i phurrng

trinh r''

-

7r'

_-

18

-

0.

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J. +

J.

_-2+

rl.

<+

-=-

. -I

t,*,,=

-2+.r,r,

rrx,

rrr,

n

)i

e

2m ₌

_{-2+ m'}

_-t<=

m)

_-2m-s

_{= 0<+}(rn+

t)(-

-:)

= o

-

nr, ₌

-l

(lo4i) hoqc rri, _{= 3}(tmdk). VAy

m=3

0,2 5

t,0
Cho tam giirc .1 /i(

'

c6 ba gric nhr.rn ( .,1

/i

< . I

(')

n6i ticp du'irng trdn

(f)).

Hai duirng

cao

tJli

ti

('t,'cria tanr giitc ..11)('cit nhau tai tli6m

//.

l)

Chri'ng minh b6n di6m

lj.('.

/i. /,' ctng thuQc mQt du'tr'ng trdn.

I

Vd cling hinh diSn y

l)

BELAC)

BEC=90''

n )5

0,25

l)

IJ

+

b6n

didm

B.('.8.F

ctng thu6c duong

tron duong

kinh

BC

0,25

,1i

0.2 5
(

2) Chf'ng minh ilutrng

thing

O.4 vu6ng g6c r'6i itu'tmg

thing

EF. 1,0

.----l

Ta co B(.:EI: ld tir giric n6i ti6p

+

AEF ₌.4BC:

Biri

IV

3.0 diOnr

2) K6 dudng

kinh

AQ

₌

L.1QC l'u6ng tai

C

-

QAC + _{AQC =90"} 0,25

0,25

0.25

x6t (o)c6

AQC = ABC =

_-s

_.)I

dAC

+

ll.:1.'

.

l.;

l()

!r(]

_{= lo}

)

l.;1:

3)

Goi

A-

li

trung diOm cria do4n

thing

B('.

Du'dng

thlng

-4 O cdt

du'ong

thing

1i('

t4i di6m

/.

dut'ng

thing

Il-F

cit

du]ng thnng .11/ tai

di6m P. Chring minh tam

gi{c

;l l'.E- ddng d4ng vcri tam

giic

AIB

vir

du'trng

thing

A1l song song v6'i du'dng

thln

_EIP

1,0

a

'io

o,

.o

tti

4

EAO

:

HAB

(vi

ctng phgr voi ABC ₎

₌

EAP =

IAB

n

)i

0.2 5

3)

_A,APE

_ddngdang_{A,AIB (g.g)}

+

O.O^

_{rrt

AB AI'

,

^AEH

ddnsdang

LABe

(s.s)

+

#=

**O,

ru

(r)

_.

_t2y

₌

A.l

₌

o.l-

+

₌

+)

_Pt

_//

_He

_t3)

AI AQ AH

AQ

Chnng minh t{r gi6c HCQB ld hinh binh hdnh

+

H,

K.Q

thing hdng (4)
Tir (3), (4)

+

KH /l

1P.

0,25

0.2 5

_T

I

l

Cl'

L lll

+

(rl'l]

= !X)"

I

</div>
<span class='text_page_counter'>(5)</span><div class='page_container' data-page=5>

Bni

v

0,5 di€m

Cho hi6u thri'c /r = a' + b'

_-

ub. voi tt.b lh c6c s6 thuc thria

min

o"' + b' + rzb _{= 3. Tim} giri

tri

l6n

nhit

vir

gii tri

nhri

nhit

cria bi6u thrlc P.

p =(a'+6't)!

-2a2b)

-ub

=

(:

_-

ra)'

-

2a)l;'

_-

ab

₌

-.a'b' _{-Tab +}

Tacb

a' +b') _{+2ab =}

3+ab+

3+ab=

3-ab

_-

a'

+b'

> 2ab

-

ab

<1

e=-[a

,

,)

(a+b)

7

2

85r

4

+

2

+ab>-3

0,5

n

),

!.(

4[

vi

_-3+!=ob*!

,t*1

₌

222

ab+-

,7

2

SI

I

2

1

4

,7

o,h+-2

>-ql

=S-'

=

p2

85-4+zr>

p>

_r

141

l1

GTLN cua P

=2l.xity rakhi

a

=..6.r,

-

--..6

toa.

u=-Ji,

t

GTNNcua

P=1, xriyrakhi

a=b=1.

.6

0,2 5

ro\(;

t)t[.\l

10,0

u

_'cl;

,l

I _l

</div>

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