解不等式log1/2^(x+1) +log2^[(1/(6-x)] 大于log1/2^12_数学解答

定义域x+1>0,1/(6-x)>0
6-x>0
所以-1log2^[(1/(6-x)]
=-log2^(6-x)
=-lg(6-x)/lg2
=lg(6-x)/lg1/2
=log1/2^(6-x)
所以log1/2^(x+1) +log1/2^(6-x)>log1/2^12
log1/2^[(x+1)(6-x)]>log1/2^12
底数1/2<1
所以log1/2^N是减函数
所以(x+1)(6-x)<12
(x-6)(x+1)>-12
x^2-5x+6>0
(x-2)(x-3)>0
x>3,x<2
结合定义域
-1