题目内容
计算:(1)| 2 |
| 2 |
(2)先化简,再求值.
| x-5 |
| x2-9 |
| x2-2x-15 |
| x2+6x+9 |
| 1 |
| x-3 |
| 3 |
分析:(1)根据零指数幂、负指数幂、特殊角的三角函数值3个知识点进行计算即可;
(2)先算除法,再算加减,最后把x=
+2代入即可.
(2)先算除法,再算加减,最后把x=
| 3 |
解答:解:(1)原式=
(2×
-1)+1-
=
(
-1)+1-
-1
=2-
+1-
-1
=2-2
;
(2)原式=
×
+
=
+
=
,
∵x=
+2,
∴原式=
=
=
=
+1.
| 2 |
| ||
| 2 |
| 1 | ||
|
=
| 2 |
| 2 |
| 2 |
=2-
| 2 |
| 2 |
=2-2
| 2 |
(2)原式=
| x-5 |
| (x+3)(x-3) |
| (x+3)2 |
| (x+3)(x-5) |
| 1 |
| x-3 |
=
| 1 |
| x-3 |
| 1 |
| x-3 |
=
| 2 |
| x-3 |
∵x=
| 3 |
∴原式=
| 2 |
| x-3 |
| 2 | ||
|
| 2 | ||
|
| 3 |
点评:本题考查了实数的运算以及分式的化简求值,解答此题的关键是把分式化到最简,然后代值计算.
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