题目内容
计算下列各式的值:
(1)(0.25)-0.5+(
)-
-6250.25-(π-3)0+2log23
(2)log43+log83)(log32+log92).
(1)(0.25)-0.5+(
| 1 |
| 27 |
| 1 |
| 3 |
(2)log43+log83)(log32+log92).
分析:(1)化小数为分数,负指数为正指数,然后运用有理指数幂的运算性质化简求值;
(2)把对数的底数的指数取倒数拿到对数符号的前面,然后通分运算即可得到答案.
(2)把对数的底数的指数取倒数拿到对数符号的前面,然后通分运算即可得到答案.
解答:解:(1)(0.25)-0.5+(
)-
-6250.25-(π-3)0+2log23
=
+(33)
-(54)
-1+3
=2+3-5-1+3=2;
(2)(log43+log83)(log32+log92)
=(
log23+
log23)(log32+
log32)
=
log23×
log32
=
.
| 1 |
| 27 |
| 1 |
| 3 |
=
| 1 |
| 0.5 |
| 1 |
| 3 |
| 1 |
| 4 |
=2+3-5-1+3=2;
(2)(log43+log83)(log32+log92)
=(
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2 |
=
| 5 |
| 6 |
| 3 |
| 2 |
=
| 5 |
| 4 |
点评:本题考查了对数的运算性质,考查了有理指数幂的化简与求值,是基础的计算题.
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