Log1/2(x+1)+log2(1/(6-x))>log1/2(12)
Log1/2(x+1)+log2(1/(6-x))>log1/2(12)
数学人气:468 ℃时间:2020-02-03 21:16:00
优质解答
首先x+1>0,6-x>0
所以-1Log1/2(x+1)+log2(1/(6-x))>log1/2(12)
即Log1/2(x+1)+Log1/2(6-x)>log1/2(12)
即Log1/2(x+1)(6-x)-log1/2(12)>0
即Log1/2(x+1)(6-x)/12>0
即(x+1)(6-x)/12<1
得x属于R
所以-1
我来回答
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