题目内容
化简
(1)(
-4)÷
(2)1-
÷(
-
).
(1)(
| x2+4 |
| x |
| x2-4 |
| x2+2x |
(2)1-
| a-1 |
| a |
| a |
| a+2 |
| 1 |
| a2+2a |
分析:(1)先把括号内通分和除法运算转化为乘法运算,再把各分子和分母因式分解得到原式=
•
,然后约分即可;
(2)先把括号内通分得到原式=1-
÷
,再除法运算转化为乘法运算得到1-
•
,约分后得到1-
,接着再进行通分后进行分式的减法运算.
| x2+4-4x |
| x |
| x(x+2) |
| (x-2)(x+2) |
(2)先把括号内通分得到原式=1-
| a-1 |
| a |
| a2-1 |
| a(a+2) |
| a-1 |
| a |
| a(a+2) |
| (a-1)(a+1) |
| a+2 |
| a+1 |
解答:解:(1)原式=
•
=
•
=x-2;
(2)原式=1-
÷[
-
]
=1-
÷
=1-
•
=1-
=
=-
.
| x2+4-4x |
| x |
| x(x+2) |
| (x-2)(x+2) |
=
| (x-2) 2 |
| x |
| x(x+2) |
| (x-2)(x+2) |
=x-2;
(2)原式=1-
| a-1 |
| a |
| a |
| a+2 |
| 1 |
| a(a+2) |
=1-
| a-1 |
| a |
| a2-1 |
| a(a+2) |
=1-
| a-1 |
| a |
| a(a+2) |
| (a-1)(a+1) |
=1-
| a+2 |
| a+1 |
=
| a+1-(a+2) |
| a+1 |
=-
| 1 |
| a+1 |
点评:本题考查了分式的混合运算:先进行分式的乘除运算(即把分式的分子或分母因式分解,然后约分),再进行分式的加减运算(异分母通过通分化为同分母);有括号先算括号.
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