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

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 (1)x(x-1)>0 f[x(x-1/2)]0 x<0或x>1/2所以(1-√17)/40=f(1) f(x)在0，＋∞）时是增函数所以 -x(x-1/2)>1x²-(1/2)x+1<0无解 综上解集为{x|(1-√17)/4