题目内容
计算
(1)
-x+2
(2)
÷
(3)1-
÷
(4)(1+
)÷
.
(1)
| 4 |
| x-2 |
(2)
| x2-9 |
| x2-1 |
| 3-x |
| x2+x |
(3)1-
| a-b |
| a+2b |
| a2-b2 |
| a2+4ab+4b2 |
(4)(1+
| 1 |
| x-1 |
| x |
| x2-1 |
分析:(1)先把各式通分,再根据同分母的分式相加减的法则进行计算即可;
(2)根据分式的除法法则进行计算即可;
(3)先算除法,再算减法即可;
(4)先算括号的,再算除法即可.
(2)根据分式的除法法则进行计算即可;
(3)先算除法,再算减法即可;
(4)先算括号的,再算除法即可.
解答:解:(1)原式=
-
=
=
;
(2)原式=
×
=
;
(3)原式=1-
×
=1-
=-
;
(4)原式=
×
=x+1.
| 4 |
| x-2 |
| (x-2)2 |
| x-2 |
=
| 4-(x-2)2 |
| x-2 |
=
| x(4-x) |
| x-2 |
(2)原式=
| (x-3)(x+3) |
| (x+1)(x-1) |
| x(x+1) |
| -(x-3) |
=
| x+3 |
| 1-x |
(3)原式=1-
| a-b |
| a+2b |
| (a+2b)2 |
| (a+b)(a-b) |
=1-
| a+2b |
| a+b |
=-
| b |
| a+b |
(4)原式=
| x |
| x-1 |
| (x+1)(x-1) |
| x |
=x+1.
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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