题目内容
计算:(1)
| x2+4 |
| x2-4 |
| 2 |
| x-2 |
(2)(a-
| a |
| a+1 |
| a2-2a |
| a2-4 |
| a+1 |
| a |
分析:(1)先对x2-4分解因式,再通分化简;
(2)先算括号里式子,再进行因式分解,最后把除法转化为乘法运算,进行分式的约分化简.
(2)先算括号里式子,再进行因式分解,最后把除法转化为乘法运算,进行分式的约分化简.
解答:解:(1)
-
=
-
=
=
=
;
(2)(a-
)÷
•
=
•
•
=a+2.
| x2+4 |
| x2-4 |
| 2 |
| x-2 |
=
| x2+4 |
| (x+2)(x-2) |
| 2(x+2) |
| (x+2)(x-2) |
=
| x2-2x |
| (x+2)(x-2) |
=
| x(x-2) |
| (x+2)(x-2) |
=
| x |
| x+2 |
(2)(a-
| a |
| a+1 |
| a2-2a |
| a2-4 |
| a+1 |
| a |
=
| a2+a-a |
| a+1 |
| (a+2)(a-2) |
| a(a-2) |
| a+1 |
| a |
=a+2.
点评:当整式与分式相加减时,一般可以把整式看作分母为1的分式,与其它分式进行通分运算.分式的化简一定要化到最简才行.
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