题目内容
计算
(1)
-
+2;
(2)
-
•
.
(1)
| 7 |
| 2a2+6a |
| 11-2a2 |
| 9-a2 |
(2)
| 1 |
| x+1 |
| x+3 |
| x2-1 |
| x2-2x+1 |
| x2+4x+3 |
分析:(1)首先将原式的分子分母能因式分解的分解因式,进而将后两项先通分运算,进而将再通分运算;
(2)首先将原式的分子分母能因式分解的分解因式,进而化简,再通分得出即可.
(2)首先将原式的分子分母能因式分解的分解因式,进而化简,再通分得出即可.
解答:解:(1)
-
+2
=
-(
-2)
=
-[
-
]
=
-
=
+
=
;
(2)
-
•
=
-
×
,
=
-
,
=
-
,
=
.
| 7 |
| 2a2+6a |
| 11-2a2 |
| 9-a2 |
=
| 7 |
| 2a(a+3) |
| 11-2a2 |
| (3-a)(3+a) |
=
| 7 |
| 2a(a+3) |
| 11-2a2 |
| (3-a)(3+a) |
| 2(3-a)(3+a) |
| (3-a)(3+a) |
=
| 7 |
| 2a(a+3) |
| -7 |
| (3-a)(3+a) |
=
| 7(3-a) |
| 2a(3+a)(3-a) |
| 7a |
| 2a(3+a)(3-a) |
=
| 21 |
| 2a(3-a)(3+a) |
(2)
| 1 |
| x+1 |
| x+3 |
| x2-1 |
| x2-2x+1 |
| x2+4x+3 |
=
| 1 |
| 1+x |
| x+3 |
| (x+1)(x-1) |
| (x-1)2 |
| (x+1)(x+3) |
=
| 1 |
| x+1 |
| x-1 |
| (x+1)2 |
=
| x+1 |
| (x+1)2 |
| x-1 |
| (x+1)2 |
=
| 2 |
| (x+1)2 |
点评:此题主要考查了分式的混合运算,正确进行分式的加减运算是解题关键.
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