Question

已知y²+3y-1=0，(y^4)/ (y^8-3y^4+1)的值。

要有过程解答。

Answer 1

y^2+3y-1=0
y+3-1/y=0
y-1/y=-3
(y-1/y)^2=9
y^2+1/y^2-2=9
y^2+1/y^2=11
(y^2+1/y^2)^2=121
y^4+1/y^4+2=121
y^4+1/y^4=119

y^4/(y^8-3y^4+1)
=1/(y^4-3+1/y^4)
=1/(y^4+1/y^4-3)
=1/(119-3)
=1/116

Answer 2

y^2+3y-1=0
y+3-1\y=0
y+1\y=-3
(y-1\y^2)=9
y^+1\y^2+2=9
y^2+1\y^2=7
(1\y^2+y^2)^2=1\y^4+2+y^4=49
1\y^4+y^4=47

y^4\(y^8-3y^4+1)
=1\(y^4-3+1/y^4)
=1\(y^4+1/y^4-3)
=1\y^4-3+1\y^4
=1\47-3
=1\44