f(x)=ax^2+x-a>1
ax^2+x-a-1>0
[ax+a+1][x-1]>0
a(x+(a+1)/a)(x-1)>0
(i)a>0,解是x>1或x1
( iii)-1/2
高中數(shù)學(xué)已知函數(shù)f(x)=ax^2+x--a.解不等式f（x)>1
高中數(shù)學(xué)已知函數(shù)f(x)=ax^2+x--a.解不等式f（x)>1
數(shù)學(xué)人氣：461 ℃時間：2020-06-19 12:01:08
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