题目内容
计算:
(1)
-
+
(2)(
)3÷(
)2
(3)
•
(4)
÷(
-x-2).
(1)
| 2a+5 |
| 2(a+1) |
| a-1 |
| 2(a+1) |
| 2a-3 |
| 2(a+1) |
(2)(
| a2b |
| -c2 |
| -a3 |
| c |
(3)
| a2+5a+4 |
| a2+a-6 |
| a2-6a+8 |
| a2-16 |
(4)
| 2x-6 |
| x-2 |
| 5 |
| x-2 |
分析:(1)分母不变.把分子相加减即可;
(2)、(4)根据分式的除法法则进行计算即可;
(3)根据分式的乘法法则进行计算即可.
(2)、(4)根据分式的除法法则进行计算即可;
(3)根据分式的乘法法则进行计算即可.
解答:解:(1)原式=
=
=
;
(2)原式=
×
=-
;
(3)原式=
•
=
;
(4)原式=
÷
=
•
=-
.
| 2a+5-a+1+2a-3 |
| 2(a+1) |
| 3(a+1) |
| 2(a+1) |
| 3 |
| 2 |
(2)原式=
| a6b3 |
| -c6 |
| c2 |
| a6 |
| b3 |
| c4 |
(3)原式=
| (a+1)(a+4) |
| (a-2)(a-3) |
| (a-2)(a-4) |
| (a+4)(a-4) |
| a+1 |
| a+3 |
(4)原式=
| 2(x-3) |
| x-2 |
| -(x+3)(x-2) |
| x-2 |
=
| 2(x-3) |
| x-2 |
| x-2 |
| -(x+3)(x-3) |
=-
| 2 |
| x+3 |
点评:本题考查的是分式的混合运算,熟知分式混合运算的法则是解答此题的关键.
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