题目内容

先化简,再求值:(
x+3
x2-3x
-
x-1
x2-6x+9
)÷
9-x
x2-3x
,其中x=2.
解法一:原式=[
x+3
x(x-3)
-
x-1
(x-3)2
9-x
x(x-3)
,(1分)
=[
(x+3)(x-3)
x(x-3)2
-
x(x-1)
x(x-3)2
9-x
x(x-3)
,(2分)
=
x2-9-x2+x
x(x-3)2
÷
9-x
x(x-3)

=
x-9
x(x-3)2
x(x-3)
9-x
,(3分)
=-
1
x-3
.(4分)
∴当x=2时,原式=-
1
2-3
=1
.(5分)

解法二:(
x+3
x2-3x
-
x-1
x2-6x+9
)÷
9-x
x2-3x

=[
x+3
x(x-3)
-
x-1
(x-3)2
]•
x(x-3)
9-x
,(1分)
=
x+3
x(x-3)
x(x-3)
9-x
-
x-1
(x-3)2
x(x-3)
9-x
,(2分)
=
x+3
9-x
-
x(x-1)
(x-3)(9-x)
,(3分)
=-
1
x-3
.(4分)
∴当x=2时,原式=-
1
2-3
=1
.(5分)
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