数学爱好者的请进！

显示全部楼层 · 2012-8-16 13:27:14

解：因为二分之一等于二的负一次方
而指数函数2的x次方是单调递增函数
所以x -3 /x +1 ≤-1所以当x大于0时，x2+2x-3≤0（x≠0）（x+3）（x-1）≤0，-3≤x≤1，所以3≤x≤1且x≠0.当x＜0时，x -3 /x +1 ≥-1，此时x≤-3或者x≥1，所以x≤-3.综上所述，得到（负无穷，-3]U(0,1]...

千问 · 2012-8-16 13:27:14

2^(x-3)/(x+1)≤1/22^(x-3)/(x+1)≤2^(-1)2>1所以(x-3)/(x+1)≤-1(x-3)/(x+1)+1≤0(x-3+x+1)/(x+1)≤0(x-1)/(x+1)≤0(x-1)(x+1)≤0-10时，整理得，（x-1）（x+3）=1(x-3)/(x+1)≤-1(x-3)/(x+1)+1≤0(x-3+x+1)/(x+1)≤0(x-1)/(x+1)≤0(x-1)(x+1)≤0-1<x≤1...