第1个回答 2011-08-08
2log a (√(4-x))-log (√a)2≥2log a (x-1)
2log a (√(4-x))-log a4≥2log a (x-1)(4-x>0,x-1>0)
log a (4-x)-log a4≥log a (x-1)^2(1<x<4)
loga[(4-x)/4]≥log a (x-1)^2(1<x<4)
a>1时,[(4-x)/4]≥(x-1)^2(1<x<4),解得1<x≤7/4
a<1时,[(4-x)/4]≤(x-1)^2(1<x<4),解得7/4≤x<4
所以原不等式的解集是(1,4)
第2个回答 2011-08-08
原式即证:2*1/2loga (4-x)-loga 4>=loga (x-1)^2 (1<x<4);
即 loga (4-x)>=loga 4(x-1)^2 (1<x<4);
当a>1时,有4-x>=4(x-1)^2,4x^2-7x<=0,0<=x<=7/4,联合1<x<4,有1<x<7/4;
当0<a<1时,有4-x<=4(x-1)^2,4x^2-7x>=0,x>=7/4或者x<=0,联合1<x<4,有7/4<=x<4;
第3个回答 2011-08-08
你的不等式等价于 (4-x)/4 >= (x-1)^2
7/4 <=x<=4