问题 解答题
计算:
(1)loga2+loga
1
2
(a>0且a≠1);
(2)(2a
2
3
b
1
2
)(-6a
1
2
b
1
3
)÷(-3a
1
6
b
5
6
)

(3)
a
1
2
a
1
2
a 
答案

(1)原式=loga(2×

1
2
)=loga1=0.

(2)原式=[2×(-6)÷3]•a(

2
3
+
1
2
-
1
6
)•b(
1
2
+
1
3
-
5
6
)
=4a.

(3)原式=

a
1
2
a
1
2
×a
1
2
=
a
1
2
a
=
a
1
2
a
1
2
=
a

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