y=f(x)(x≠0)是奇函数,且当x属于（0,＋∞）时是增函数,若f(1)=0,求不等式f[x(x-1\2)]＜0解集
问题描述：
y=f(x)(x≠0)是奇函数,且当x属于（0,＋∞）时是增函数,若f(1)=0,求不等式f[x(x-1\2)]＜0解集
答
f[x(x-1/2)]0
f[x(x-1/2)]>f(1)
f(x)在0,＋∞）时是增函数
所以 x(x-1/2)>1
x²-(1/2)x-1>0
2x²-x-2>0
x>(1+√17)/4或x0
f[x(x-1/2)]>0
-f[-x(x-1/2)]>0
f[-x(x-1/2)]答案是1\2＜1＋√7／4或1-√7／4＜x＜0晕菜，我按大于0做了，抱歉f[x(x-1/2)]0 f[x(x-1/2)]