题目内容
计算或化简:(1)(
| 1 |
| 2 |
| 3 |
| 2 |
| x-1 |
| 4 |
| x2-1 |
分析:(1)本题需先根据负正数指数幂、零指数幂、特殊角的三角函数值、绝对值的运算,即可求出答案.
(2)本题需先根据分式的加减运算分别进行计算,即可求出答案.
(2)本题需先根据分式的加减运算分别进行计算,即可求出答案.
解答:解:(1)(
)-1-(2010-
)0+4sin30°-|-2|;
=2-1+4×
-2,
=1;
(2)原式=
-
,
=
,
=
,
=
.
| 1 |
| 2 |
| 3 |
=2-1+4×
| 1 |
| 2 |
=1;
(2)原式=
| 2(x+1) |
| (x+1)(x-1) |
| 4 |
| (x+1)(x-1) |
=
| 2x-2 |
| (x+1)(x-1) |
=
| 2(x-1) |
| (x+1)(x-1) |
=
| 2 |
| x+1 |
点评:本题主要考查了分式的混合运算,在解题时要根据分式运算的顺序和法则分别进行计算是本题的关键.
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