问题
解答题
(1)计算:
(2)已知lga+lgb=21g(a-2b)求
|
答案
(1)原式=lg
+2
-(lg
-1)22
÷3 a
•a-9 2 3 2 3 a
•a-13 2 7 2
=lg
+1-lg2
-12
=0
(2)∵lga+lgb=2(lg(2-2b)
∴lg(ab)=lg(a-2b)2
∴ab=(a-2b)2即a2+4b2-5ab=0∴(
)2-5a b
+4=0a b
解之得
=1或a b
=4a b
∵a>0,b>0若
=1则a-2b<0a b
∴
=1(舍去)a b
∴
=4a b