đề thi vào lớp 10 môn toán tỉnh hà nam năm 2015-2016

UBNN
riNs
HA
NAM
or
rru
ruyEN
sINH
Lop
10
THpr
NAnr
Hec
20ts
-
z0t6
sO
GrAo
pUC
vA
DAo
rAo
M6n:
To6n
op
cuiNu
rHrlc
c6u
1
(2,0
cti€m).

Thdi gian
ldm
bdi;
t20 philt,
kh6ng
ke
thdi gianphdt
d€
2
2-rlx
a)
Gi6i phuong
trinh
5x
b)
Giei
hd phuong
trinh
Cf,u
3
(1,5
di€m).
Trong
mpt
phing
tga
d6
Oxy
cho
parabol

(p):
,-
_ii
y
=
3mx
-
3
(v6i
m ld
tham
s6).
.,
,l
'-6x-8=0.
I(.
*3)(y*
2)=7
+xy
[(**l)(v*
t)=*y*2.
a)
Rut gon
bi6u
thric
A
=
rA.
-
7Ji

+
S",60
.
b)Chobi6uthricB=^-G*
-1
(v6i
x>0
vd
x
+4).
Riit
ggn
B
Ciu
2 (1,5
tti€m).
x-4
vdtimxdC
B=1.
i,i
'1,
'
':'
vd
dubng
thdng
(d):
a)
Tirn
m

d6
ducrng
thing (d)
di
qua
di6m
A(1;
3).
b)
Xdc
dinh
c6c giStri
cua
m
9i
(d)
cet (P)
tq\:nitdi6m
ph6n
bi6t
sao
cho
t6ng2
tung
dQ
cua
hai
giao
di6m
d6

bing -10.
cho
ducrng-lidn(o)^v?.dicm
A ra-
trdn
duong
trdn.
9qi
o
ld
ti6p
tuy5n
cira
(o)
laiA,
rr6n
d
l6ydi6m
D
(D
kh6ngt*ns
v6{
A),
k6
ii6t66r
r;
.u,
r.D
(e
iJdia;,

B
kh6ng
trung
v6i
A).
;:'
a)
Chfng
-j*
ring
tf
gidc.
AOBD
n6i
ti6p.
b)
Tr6n
tia
doi
cria
tia BA
l6y
di6m
C.
Ke
bH
vuong
g6c
v6i
OC

(H
thu6c
OC).
Gqi I
ld
giao
dicm
cira
ae ;5
oo.
"ch;G
-i"r,
*"g
oH.oc
=
oI.oD.
c)
Gqi
M la.giao
di6m
cua
DH
v6icung
nho
AB
cria (O).
Chrmg
minh
ring
CM

Ia
tiep
tuydn
cua
1O).
+
\
d).Gqi
E l+ gi.uo.
diem
cria
DH
vd
CI.
Goi
F
ld
giao
di6m
thf
hai
cira
ducyng
tron
9i"g.
kinh
oD.va
dudng
trdn
ngoai

ti6p
tam gi6c
oirra.
chrmg
minh
rdng
o, E, F
thang
hang.
Ciu
5.(!1,0
Aiem1.
,.
tt,
,
@ho
x,
y
ld
c5c
s6
thuc
duong
thoa
m6n
x
+
3y
_<
l0.

,.:
rt-:
:
'
D6u
ding
thr.
rayl;*I;J",

HBT

Girim
thi
tht
nh6t:

Gi6m
thi
thrl
hai:
HT.IONG
OAX
CAr
Or-rWEN
SINH
LOP
10
THPT
riNrr
HA

NAM
NAM
HQC
20ts
_2016
Mdn:
Todn.
NQi
dung
A=2Ji
-28J,
+zsJ2
Ta
co
l':
(-3)'
-
5(-8)
=
49
>
O
tr
,,
prc6hainghiemph6nbi6t
x,
-3+7=z;
i
''*
4.

)55
I(^*3Xy
+2)-z*
[(**1)(V*
l)=*y
+2
-"
l*r+x+y+l
=
xy+2
,
,:., 4:-i
lZx
+3y
=
I
lZ*
+3v
=
I
al
€1
<f<
[x+y=1
l.:x+3y=3
'
;
lt';'.
(u
-)

l/\-L
r-r J
\-,|
lv=-1
w
Vay
h.g
phuong
trinh
c6
nghi6m
duy
nh6t
(x;
y)
:
(2;
_1).
Dtrong
thdng
(d)
di qua
A(
Phuong
trinh
hodnh
d0
Siu
-x'
=3mx

-3
<+
x2
+3mx-3
=
0 (*)
Ta
c6
L
=9m2
+12
>
0,
v6i
moi
m
n6n
phucrng
trinh (*)
c6
hai
nghidm
ph6n
bi6t.
Do
d6
ducrng
thing
(d)
vd

parabol
(p)
c6t
nhau
tqihaidiCm
(x,,y,)
vd
(*r,yr).
fneo
dinh
ly
Vi-6t
ta
c6
xr
+
x2
=
-3m
i
Xr.Xz
-
_3
.
Theo
bdi
ra
ta
c6 yr
* yz=

-10
o
_^i
_i
=
-10
e
(",
+
*rl'
-
2xrx,
Dod6
9m2
+6=10<+m
-*?
DA
vd
DB
te
c6c
ti6p
tuy-6n@lO;,ren
6BDE
Xdt
tf gi5c
AOB,
:U'
d"t5
*

OD
=
1g00,
md
hai
g6c
ndy
6
vi
tri
aOi
aien
n6n
tir gi6c
AOBpail,i
tiep.
b)
Theo
tinh
clr6j.'hei
figp
^
gi6c
cria
D
.
Do
d6
tam
gi6c

ABD
cdn
taiD
c6
Do
re
ducrng
phdn
gi6c
n€n
ddng
thcyi
ld
ducrng
trung
tryc
,,,
ic)
usv.r.
"xdt
AoIC
vd
AOI{D
c6
5id
=
ofu
=
900;
chung

f6D
n6n
AOIC.,,AOFID(g.g)
OI
OC

=OH.OC=OI.OD.
(t)
OH
OD
Xdt
tam giSc
AOD
vu6ng
tai
A
c6
AI
ld
(2)
Ma
OM
=
OA
(ld
b6n
kinh
(O).
(3)
Tt

(1),
(2)
vd(3)
suy
ra
OM2
=
OH.OC
=
OM
=
OC
OH
OM.
Xdt
AOHM
vd
AOMC
AOHM.,,AOMC(c.g.c)
.
+ OMa
-
6id
=
900
n6n
CM ld
tiiip
tuy6n
cta

(O).
c6
chung
Moa;
gIuI
=
oc
,c,
OH
OM
Do
OMC
=
OIC
=
900
n6n
tri
gi6c
OIMC
n6i
ti6p
ducrng
trdn
ducrng
GF
OC.
,o.,nia-
l.
Dudng

trdn
ngopi
tiop
tam gi6c
cIM
ld
duong
trdn
ducrng
kinh
oc.
^
1
suy
ra
OFC
=
900.
,]:1|||
,
:.|
Mpt
kh6c
ta
c6
oFD
=
900.
Nhu
v6y

G-,
m
rc uisJv
rabadiem
C,
F,
''
.'t
"a
D
thdng
hdng.
"
'
a
.:.
Xdt
tam gi6c
ocD
c6.ba
dudng
cao
cH,
oq
di,md
c6
E ld giao
di6m
cH,
DI

n6n
ba di6m
O, E,'F
thing
hdng.
C6ch
1.Ap
dung
bgjtoet,i.
1
1
l1
r+ t-+a)l^:/
, ;.xe++x>3.
(l)
Vx
Vx
VVxJx Jx
2727^f2?n2.27
_++_L*3v
>
3l_L.4.5
CQng
cdc
bdp
d.tngthuc
(
1) vd (2)
tadugc
/\

,[*
-#)+(x
+:v)
>30
e
r[#
.
#)>30-(x+3y)
=
2e
x
+3:!r,'<10).
127
<]-/-+-/->10.
D6u
ding
thtic
xay
ra
khi
vd
chi
*'
t;=;
C6ch
2.
Ap
dUng
b6t
tang

thric
Bunhiacopski
,
ta
c6
I
*
27
_
1
_3:27
_-
(t+e)'
1oo
E-E=G-
3\EzG;m=F+\F'
(r)
(vi
(,
E
#$y)
=
(r'
+
r'X*
+
3y)
<
100
=

J*
*3$y<
10.
(2)
Tir
(1)
va
(2)
suy
ra
** +>
10.
J*
J:y
-^
ryyrakhivdchithi
I^=1
[Y=3