题目内容
计算:
(1)
+
;
(2)
÷(
-2).
(1)
| m2 |
| m-2 |
| 4 |
| 2-m |
(2)
| 3-x |
| 2x-4 |
| 2 |
| x-2 |
考点:分式的混合运算
专题:
分析:(1)首先把式子进行通分相减,然后对结果进行化简即可;
(2)首先把括号内的式子进行通分相减,把除法转化为乘法运算,最后进行乘法计算即可.
(2)首先把括号内的式子进行通分相减,把除法转化为乘法运算,最后进行乘法计算即可.
解答:解:(1)原式=
-
=
=
=m+2;
(2)原式=
÷
=
•
=
.
| m2 |
| m-2 |
| 4 |
| m-2 |
=
| m2-4 |
| m-2 |
=
| (m+2)(m-2) |
| m-2 |
=m+2;
(2)原式=
| 3-x |
| 2(x-2) |
| 2-2(x-2) |
| x-2 |
=
| 3-x |
| 2(x-2) |
| x-2 |
| 2(3-x) |
=
| 1 |
| 4 |
点评:本题主要考查分式的混合运算,通分、因式分解和约分是解答的关键.
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