题目内容
计算:(1)(
×
)6+(
)
-4(
)-
-
×80.25-(-2005)0
(2)log2.56.25+lg
+ln(e
)+log2(log216)
3 | 2 |
3 |
2
|
4 |
3 |
16 |
49 |
1 |
2 |
4 | 2 |
(2)log2.56.25+lg
1 |
100 |
e |
分析:(1)利用有理数指数幂的运算法则,把原式等价转化为(2
×3
)6+(2
×2
)
-4×
-2
×2
-1,由此能求出结果.
(2)利用对数的运算性质把原式等价转化2-2+
+log24,由此能求出结果.
1 |
3 |
1 |
2 |
1 |
2 |
1 |
4 |
4 |
3 |
7 |
4 |
1 |
4 |
3 |
4 |
(2)利用对数的运算性质把原式等价转化2-2+
3 |
2 |
解答:解:(1)原式=(2
×3
)6+(2
×2
)
-4×
-2
×2
-1
=22×33+2-7-2-1
=100.
(2)原式=2-2+
+log24
=
+2
=
.
1 |
3 |
1 |
2 |
1 |
2 |
1 |
4 |
4 |
3 |
7 |
4 |
1 |
4 |
3 |
4 |
=22×33+2-7-2-1
=100.
(2)原式=2-2+
3 |
2 |
=
3 |
2 |
=
7 |
2 |
点评:本题考查有理数指数幂的运算法则和对数的运算性质,是基础题.解题时要认真审题,仔细解答.
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