Log1/2(x+1)+log2(1/(6-x))>log1/2(12)
问题描述:
Log1/2(x+1)+log2(1/(6-x))>log1/2(12)
答
首先x+1>0,6-x>0
所以-1
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得x属于R
所以-1