Đề thi và đáp án Toán vòng 1 Diễn Châu 2014-2015

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9
NAM
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20r4_20t5
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('Itroi
gian
lArn
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150 phut)
Ciu

1.
(6
di6rn)
I
.
'l'irrh
gid
tri
cua
bi0u thuc:
A
:
*
Jl
_ qJl
;
2.
(ltro
bi0u
tliu.c:
I) =.
.2' r =-
4

L-
x-2
x'-5x+6
x-3
a)
l'im

diOu
ki€n
cua
x
clO
t' co
nghTa
vd
rut
gon
bi6u
thric
p.
b)
f im
tAt
cd
cdc gi|tri
cua
x
nguy0n
d0
p
nguyOn.
CAu 2.
(4
di6m)
2.
'fin
nghie

rn
nguyOn
t6
oria phucrng
trinh:
x2
-
2y2
-
1
CAu
3.
(a
cliOrn)
I
cho
bi6u
thrl'c
A
=
x
-zJi
+3y-2Ji
+1.
Tim gi6
trf
nho
nhAt
md
A c6

th€
dat
0u'0c.
2.
(''ho
x,
Y,
Z ld
ba
s6
du'cmg
vd
x
+
y
+
z:4.
Chri.ng
rninl,
ring
x
-r
y
>
xyz.
C'iiu
4. (4
di0n)
l.
Clro

hlnh vu6ng
ABCD
c6
c4nh
bing
a.
Lay di€m
M
tiy
ii
tr0n
ducrng
chdo
AC,
kd
ME
r
AB,
MF
r
BC.
X6c
dinh
vi tri
cria
M tr6n
cluong
chio
AC
d6 dien

tich
A I)EF
nho
nhAt,
tim gi6
trf
d6.
2.
Ctro
hinh
binh
hdnh
ABCD
vdi
AC
>
DB.
Goi
E vd
F
lAn
lucrt
ld chdn
duong
vu6ng goc
lcd
tu
C d6n
cdc
dudrng

thing
AB vd
AD.
Chu'ng
rninh
ring:
AB.All
+
AD.AF
:
AC'
C'iu
5. (2
diem)
Cho
ba s6
a,
b,
c
thoa
mdn
a
+
b
+
c
:
0 vd
m6i
s6 cl6

ldn
hon
-
1 vd
nho
hcrn
1.
/ir
I
J
Chri'ng
rninh
rdng:
a'
+
b2
+
c'
<2.
HUONG
nAN
CUAivr
I\{ON
rOAn
Cffu.
Dip
6n
Bi€u
*.
I

olem
I
a-2Ji
+
7
-4Jj
3-2",6+1+
1.A=
a-all+3
1,0
:
.6-r+2 6:1
/
-
r2
IVt
-
t,1
1,0
2. a)P

2x
4
1 2x
A
T
l-
,dlk:
x+2;
x+3

x-3'
:_
x-2
x'-5x+6
x-3 x-2
+
(x-
z)(x-:)
0,5
2x(x4)+a-(r-z)
_
(x
-
z)(x
-:)
2x2
-7
x +6
(x
-2)(x
4)
0,5
(x
-z)(zx -z)
_2x-3
x-3
(x
-
z)(x
-:)

0,5
)lr-7 ?
b)P:
-'"
i
:
:2+
"
'
x-3
x-3
0,5
DO P nguyOn thi
x
-
3
phdi
ld
u6c cua
3
1,0
U(:):
{t;-t;:;-:}
Vdi x-3
:
I
:)
x
:
4; Vdi

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:
-l
:>
x: 2(loai);
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-
3
:
3:) X: 6; v6ix-3
:
-
3
:>
x
:
0. 1,0
Vqy
cdc
gi6
tri cdn tim ld x:
6; x: 4; x: 0.
2
1. x'
+y' +
22
:xy
+
yz*
zx e

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y)'
*(y-t)' +(r-")'
:0
0,5
<>
x:
y
:
z.
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do:
x2014
+
y'o'o
+220t4
-
3rzot+
-
3
20rs
0,5
->X2014 -32ott:)X:3.
0,5
Vfly hq c6 nghiQm: x
:
y
:
z:
3.

0,5
2.Tac6 x2
-2y'=l
(l)
x
-
-l
=
/1)-
_ _-r
0,25
e ("+1)(x
-t)=2y'
0,25
:t
(r+1)(x-1):
2
(x)
0,25
Md
x
+
1 vd x
-
I cins tinh chAt chdn, 16 do d6:
0,25
(*):>x+li2vitx-l:2 0,25
:>zyri
4
0,25

:>y2i2
Md2 nguyOn t6:>
y
i 2
0,25
y
lai ld sd nguyOn t6
:)
y
:2
Tir
(1):)
x:
3
0,25
Vay: nghiQm
nguy0n td cria
(l)
ld X:
3,
y
:2.
-
J
1.A: r-2,[ry+3y-2Ji+1.Di0ukiOn
x>0 vd
y>0
d0
Ji
vd

Ji
"6nghia.
+Vdix>0vdy>0,khid6:
A
:
x-2.,1i
+
y+2y-2J;
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:
(J; D)'
+t-2H;
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2^[i +2y
0.5
:
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J,
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-r [-,
+2v +:
:
:
(G
-
Ji
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-:
:)A
,-;
vay
A
datgi6trinh6
nrrat
wri
zJv
-
1:0
vd
J;-Jr-1=0
t9
eV:

x=-
'
4' 4
1
Dod6MinA:-1
L
+
Vdi x:
0 vdy
>
0,
khi d6 A
:
r-zJi

+
y+2v-zJi+1
:
3y
+
1
:)
A
khdngt6n
t4i
cpc
tri.
+V6ix>0vdy:0,khid6A:
*-2'[*y+y+2y-zJi+l
:x
-
zJi
+
1:
.1
(.'6-r)'
>
0v6imgi
x>
0. A:
Q e
Jt-r=0
ex:
1
Z.Talu6nc6:

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)0
:)
*t
*y'+2xy>4xy:>
("
*y)'2a*y
(l)
Ap
dung
(1),
ta co
thO
vi6t: <+
[('*y)
*"f'
>a(x+y)z
Vi x
+
y
+
z:4
n6n: (+)'
>
a(x+
y)z
:>
16
(**y)>a(**y)'
,

(vi
x>
0,
y>
0):t
16(x+
t)>rcty,
(theo
(1))
Vflyx*y>-ry"
Dfr"-_'"ay
*khi
x
*
y: zngYfia
ld
x:
y:
I
vd z:2
a ndn
BE.BF
lcrn
nhAt
e
BE
:
BF
L(
",

-4-\:10,
2[
4)
8
i-n€
thAy
S
MEM
:
S
*,
(cirng
chiOu
cao,
cirng
canh
d6y).
Tucrngtu:
S-rr:So
+
S-rr:S-rt*S-rt*S**
:
Sorrc
-
S*r,
:
IV'
-
BE-BF)
S

-u,
dat
gi|tri nho
nh
o
Ut.Ut
lcrn
nhat
(1)
e
M
ld
trung
di6m
cua
AC
vd
S
-rr:
0,25
0,25
0,25
0,25
0,25
0,25
0,5
LK1BG
IAC,tac6:
AACE
-

AABG
AC
AB,
:)
-
=-
nav
AC.AG
:
AE.AB
(1)
AE
AG
-
-
Vi AF//CB
ndn
6CB
=
EAF
:>
LCBG
-
A
ACF
AC
BC
_>_
AF
CG

Hay
AC.CG:
AF.BC
(2)
0,25
0,5
0,25
0,5
0,25
025
c0ng
(1)
vo-i
(2)
theo
tung
v0,
v6i
chri
y
BC =
AD
vd
AC
>
BD
non
diom
G
^

L^

^
^n
A E r AT\ AD
-
LC2
0,25
'l'ua+
bl c:0
xiry
ra
ba trudng
trop
sau:
l) NOu
b
-
0 thi
a2+
b2+
c'
:
a'
+
c'<
I
+
|
=2.

-
|
u
I
v
_4
ru \I 1_L_2.
2)
NOu
I
<
a<
0
<b
< c
<
l,
thi
0
<b*c=-
a<
1.
n€naz+b2
*c2 <
(-u)+(b+c)<
1+
l:2.
3)NOu
I
:;ab.O.c

(
l,thi0
<-a<
l,
0
<-b<
1 vd0
<c<-(a+b)
<
l,
-
Ildi
hinh,
khong co
hinh
vo
ho4c
v0
hinh
sai
kh6ng
chAm
bdi hinh.
u
thi