Question

已知：x²+3x+1=0,求下列各式的值：（1）x+(1\x);(2)x²\x的四次方+1

Answer 1

x^2+3x+1=0 -> x+1/x+3=0

(1) x+1/x=-3

(2) x^2/(x^4+1)=1/(x^2+1/x^2)

由(1)得 (x+1/x)^2=(-3)^2=9 -> x^2+1/x^2+2=9 -> x^2+1/x^2=7

所以x^2/(x^4+1)=1/7