®Ò kiÓm tra
M«n :To¸n
Thêi gian :45 phót
C©u 1 (5®)
Gi¶i c¸c pt sau :
a)
2
1 3 4
1 1 1
x x
x x x
+
+ =
− + −
b)
4 5 4x x+ = −
c)
2 1 1x x+ = −
C©u 2 :(4®)
Cho ph¬ng tr×nh :x
2
-2(m+1)x+m
2
-4=0
a)Gi¶i pt víi m=1
b)X¸c ®Þnh m ®Ó pt cã hai nghiÖm mµ x
1
.x
2
=0.TÝnh c¸c nghiÖm ®ã .
c)T×m m ®Ó pt cã nghiÖm
C©u 3:(1®):
Gi¶i hÖ pt :
5
2 3 14
x y
x y
+ =
+ =
§¸p ¸n
ý c©u §¸p ¸n
®iÓm
1 a)®k :x
1≠ ±
( )
( )
2 2
2
( 1) 1
3 4
1
1 1
1 21
2
5 0
1 21
2
x x x
x
x x
x
x x
x
+ + −
+
⇒ =
− −
−
=
⇒ − − = ⇒
+
=
b)
dk :x
5
4
−
≥
( )
( )
2
2
4
4
4
2 11
1
12 11 0
4 5 4
11
x
x
x
x
x
x x
x x
x
≥
≥
≥
⇔ ⇔ ⇔ ⇔ =
=
− + =
+ = −
=
c)
1 0 1
2 1 1 2
2 1 1 0
x x
vn
x x x
x x x
− ≥ ≥
⇔ ⇔ ⇔
+ = − = −
+ = − =
0,5®
0,5d
1d
0,5d
1d
1,5
2
3
a)khi m=1
pt trë thµnh :x
2
-4x-3=0
cã
,
7∆ =
x=2
7±
b)dk :
/
2
1 2
5
2 5 0
0
2
2
. 0
4 0
2
m
m
m
x x
m
m
−
+ ≥
≥
∆ ≥
⇔ ⇔ ⇔ = ±
=
− =
= ±
+)m=2
2
0
2 0
2
x
x x
x
=
⇒ + = ⇔
= −
+)m=-2
2
0
6 0
6
x
x x
x
=
⇒ − = ⇔
=
c)®k :
( )
2
/ 2
5
0 1 4 0 2 5 0
2
m m m m
−
∆ ≥ ⇔ + − + ≥ ⇔ + ≥ ⇔ ≥
5 1
2 3 14 4
x y x
x y y
+ = =
⇔
+ = =
1,5d
1d
0,25d
0,25d
1d
1d