y=f(x)(x≠0)是奇函數(shù),且當x屬于（0,＋∞）時是增函數(shù),若f(1)=0,求不等式f[x(x-1\2)]＜0解集

數(shù)學人氣：918 ℃時間：2020-01-29 14:21:59

優(yōu)質(zhì)解答

f[x(x-1/2)]0
f[x(x-1/2)]>f(1)
f(x)在0,＋∞）時是增函數(shù)
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暈菜，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，＋∞）時是增函數(shù)所以 -x(x-1/2)>1x²-(1/2)x+1<0無解綜上解集為{x|(1-√17)/4