计算: (1)(25/9)0.5+(27/64)−2/3+(0.1)−2−3π0; (2)lg1/2−lg5/8+lg12.5−log89•log278.
问题描述:
计算:
(1)(
)0.5+(25 9
)−27 64
+(0.1)−2−3π0;2 3
(2)lg
−lg1 2
+lg12.5−log89•log278. 5 8
答
(1)原式=[(
)2]0.5+[(5 3
)3]−3 4
+(10-1)-2-3×12 3
=
+5 3
+100−316 9
=100
.4 9
(2)原式=lg(
×1 2
×8 5
)-25 2
•lg9 lg8
lg8 lg27
=lg10-
lg32
lg33
=1-
2 3
=
.1 3