题目内容
(2013•泰州)(1)计算:(
)-1+|3tan30°-1|-(π-3)0;
(2)先化简,再求值:
÷(x+2-
),其中x=
-3.
1 |
2 |
(2)先化简,再求值:
x-3 |
x-2 |
5 |
x-2 |
5 |
分析:(1)根据负指数幂、特殊角的三角函数值、0指数幂的定义解答即可;
(2)将括号内的部分通分,再将除法转化为乘法,然后代入求值.
(2)将括号内的部分通分,再将除法转化为乘法,然后代入求值.
解答:解:(1)原式=
+|3×
-1|-1
=2+|
-1|-1
=1+
-1
=
;
(2)原式=
÷(
)
=
÷
=
•
=
.
当x=
-3时,
原式=
=
=
.
1 | ||
|
| ||
3 |
=2+|
3 |
=1+
3 |
=
3 |
(2)原式=
x-3 |
x-2 |
x2-4-5 |
x-2 |
=
x-3 |
x-2 |
(x-3)(x+3) |
x-2 |
=
x-3 |
x-2 |
x-2 |
(x-3)(x+3) |
=
1 |
x+3 |
当x=
5 |
原式=
1 | ||
|
1 | ||
|
| ||
5 |
点评:(1)本题考查了实数的运算,涉及负指数幂、特殊角的三角函数值、0指数幂的定义,是一道简单的杂烩题;
(2)本题考查了分式的化简求值,熟悉通分、约分和分式的加减是解题的关键.
(2)本题考查了分式的化简求值,熟悉通分、约分和分式的加减是解题的关键.
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