问题 解答题
(1)计算:
1
2
lg2+
(lg
2
)
2
-lg2+1
-
3
a9
a-3
÷
3
a13
a7

(2)已知lga+lgb=21g(a-2b)
a
b
的值.
答案

(1)原式=lg

2
+
(lg
2
-1)
2
-
3a
9
2
a-
3
2
÷
3a
13
2
a-
7
2

=lg

2
+1-lg
2
-1

=0

(2)∵lga+lgb=2(lg(2-2b)

∴lg(ab)=lg(a-2b)2

∴ab=(a-2b)2即a2+4b2-5ab=0∴(

a
b
)2-5
a
b
+4=0

解之得

a
b
=1或
a
b
=4

a>0,b>0若

a
b
=1则a-2b<0

a
b
=1(舍去)

a
b
=4

单项选择题
单项选择题