Question

已知y²+3y-1=0,求y^4／(y^8-3y^4+1)的值.

Answer 1

y^4/(y8-3y4+1)

＝1/y^4-3+1/y^4=1/(y^2+1/y^2)^2-5=1/[(y-1/y)^2+2]^2-5

由y2+3y-1=0 两边除以y 代入上式

则 原式＝[(-3)^2+2]^2+5=1/126