题目内容
计算:
(1)2 -
+
+
-
;
(2)log2.56.25+lg0.001+ln
+2log23.
(1)2 -
1 |
2 |
(-4)0 | ||
|
1 | ||
|
(1-
|
(2)log2.56.25+lg0.001+ln
e |
分析:(1)通过变分数指数幂为根式及非0数的0次幂为1进行计算;
(2)运用对数式的运算性质直接化简求值.
(2)运用对数式的运算性质直接化简求值.
解答:(1)解:2-
+
+
-
=
+
+(
+1)-1
=
+
+
+1-1
=2
.
(2)log2.56.25+lg0.001+ln
+2log23
=log2.52.52+lg10-3+lne
+2log23
=2-3+
+2log23=2log23-
.
1 |
2 |
(-4)0 | ||
|
1 | ||
|
(1-
|
=
1 | ||
|
1 | ||
|
2 |
=
| ||
2 |
| ||
2 |
2 |
=2
2 |
(2)log2.56.25+lg0.001+ln
e |
=log2.52.52+lg10-3+lne
1 |
2 |
=2-3+
1 |
2 |
1 |
2 |
点评:本题考查了有理指数幂的化简求值及对数式的运算性质,解答此题的关键是熟记有关的运算性质,是基础题.

练习册系列答案
相关题目