题目内容
化简求值:(1)
| x2-4 |
| x2-4x+4 |
| x2-1 |
| x |
| 2 |
(2)(
| 3x |
| x-1 |
| x |
| x+1 |
| x2-1 |
| x |
(3)(
| x-y |
| x2-2xy+y2 |
| xy+y2 |
| x2-y2 |
| xy |
| y-1 |
| 1 | ||
2-
|
| 1 | ||
2+
|
分析:(1)分子分母分别分解因式后可发现都含有x-2这一因式,约分化简后代入x的值计算即可解答;
(2)首先将括号里面的两个因式通分,再与后面的因式约分,化成最简分式,再代值计算即可解答;
(3)将括号里面的两个因式通分,再与后面的因式约分,化成最简分式,再代值计算即可解答.
(2)首先将括号里面的两个因式通分,再与后面的因式约分,化成最简分式,再代值计算即可解答;
(3)将括号里面的两个因式通分,再与后面的因式约分,化成最简分式,再代值计算即可解答.
解答:解:(1)原式=
•
=
,
当x=
-1时,
可得
=
=
+
.
(2)原式=
•
=3(x+1)-(x-1)=2x+4,
当x=2时,2x+4=8.
(3)原式=(
-
)•
=-
,
由x=
,y=
,
可知x=2+
,y=2-
,
代入可得
=
.
| (x+2)(x-2) |
| (x-2)2 |
| x2-1 |
| x |
=
| (x+2)(x2-1) |
| x(x-2) |
当x=
| 2 |
可得
(
| ||||||
(
|
=
| -2 | ||
5-4
|
=
| 10 |
| 7 |
8
| ||
| 7 |
(2)原式=
| 3x(x+1)-x(x-1) |
| x2-1 |
| x2-1 |
| x |
=3(x+1)-(x-1)=2x+4,
当x=2时,2x+4=8.
(3)原式=(
| 1 |
| x-y |
| y |
| x-y |
| xy |
| y-1 |
=-
| xy |
| x-y |
由x=
| 1 | ||
2-
|
| 1 | ||
2+
|
可知x=2+
| 3 |
| 3 |
代入可得
(2+
| ||||
(2+
|
| ||
| 6 |
点评:本题主要考查分式的化简求值的知识点,解答本题的关键是把两分式合并后约分为最简,本题难度一般,代值计算要仔细.
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