解不等式(x²+4x-5)(x²-2x+2)>0

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解不等式(x²+4x-5)(x²-2x+2)>0
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解不等式(x²+4x-5)(x²-2x+2)>0
解不等式(x²+4x-5)(x²-2x+2)>0

解不等式(x²+4x-5)(x²-2x+2)>0
(x^2+4x-5)(x^2-2x+2)>0
因为:x^2-2x+2=(x-1)^2+1>0
所以,原不等式等同于:x^2+4x-5>0
即:(x+5)(x-1)>0
x1

(x+5)(x-1)((x-1)^2+1)>0
(x-1)^2+1>=1
(x+5)(x-1)>0
x>1 or x<-5

第二个方程恒大于0,故只用第一个方程大于0就行了

(x²+4x-5)(x²-2x+2)>0
(x+5)(x-1)*<(x-1)^2+1)>>0
(x-1)^2+1一定大于0

(x+5)(x-1)〉0
x>1或x<-5