题目内容
计算:
(1)
÷
;
(2)
-
÷
;
(3)(1-
)÷
.
(1)
| a-1 |
| a-2 |
| a2-2a+1 |
| 2a-4 |
(2)
| 1 |
| x+1 |
| 3-x |
| x2-6x+9 |
| x2+x |
| x-3 |
(3)(1-
| a2+8 |
| a2+4a+4 |
| 4a-4 |
| a2+2a |
考点:分式的混合运算
专题:
分析:(1)利用分式的混合运算顺序求解.
(2)利用分式的混合运算顺序求解.
(3)利用分式的混合运算顺序求解.
(2)利用分式的混合运算顺序求解.
(3)利用分式的混合运算顺序求解.
解答:解:(1)
÷
=
•
,
=
.
(2)
-
÷
=
+
•
,
=
+
,
=
,
=
,
(3)(1-
)÷
=(1-
)•
,
=
•
,
=
.
| a-1 |
| a-2 |
| a2-2a+1 |
| 2a-4 |
=
| a-1 |
| a-2 |
| 2(a-2) |
| (a-1)2 |
=
| 2 |
| a-1 |
(2)
| 1 |
| x+1 |
| 3-x |
| x2-6x+9 |
| x2+x |
| x-3 |
=
| 1 |
| x+1 |
| x-3 |
| (x-3)2 |
| x-3 |
| x(x+1) |
=
| 1 |
| x+1 |
| 1 |
| x(x+1) |
=
| x+1 |
| x(x+1) |
=
| 1 |
| x |
(3)(1-
| a2+8 |
| a2+4a+4 |
| 4a-4 |
| a2+2a |
=(1-
| a2+8 |
| (a+2)2 |
| a(a+2) |
| 4(a-1) |
=
| 4(a-1) |
| (a+2)2 |
| a(a+2) |
| 4(a-1) |
=
| a |
| a+2 |
点评:本题主要考查分式的混合运算,解题的关键是熟记分式的混合运算顺序.
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