已知:x^2+3x+1=0,求x^2+x^2/1的值
问题描述：
已知:x^2+3x+1=0,求x^2+x^2/1的值
答
原式=[3(x+1)]^2-[2(x-1)]^2
=(3x+3)^2-(2x-2)^2
=(3x+3+2x-2)(3x+3-2x+2)
=(5x+1)(x+5)
1/x+1/y=(x+y)/xy=5
x+y=5xy
所以原式=[2(x+y)-5xy]/[(x+y)+2xy]
=[2(5xy)-5xy]/[(5xy)+2xy]
=5xy/7xy
=5/7
答
x^2+1=-3x
平方
x^4+2x^2+1=9x^2
x^4+1=7x^2
两边除以x^2
x^2+x^2/1=7