倒数法求值,已知x/x^-x+1=5,求x^/x的四次方+x^+1的值?

Question

倒数法求值,
已知x/x^-x+1=5,求x^/x的四次方+x^+1的值?

Answer 1

因为x/(x^2-x+1)=5,
所以(x^2-x+1)/x=1/5,即 x-1+1/x=1/5,x+1/x=6/5,
故 (x+1/x)^2 = 36/25,即 x^2 + 2 + 1/x^2 = 36/25,
所以 x^2 + 1/x^2 = 36/25 -2 = -14/25,
因此 x^2 + 1 + 1/x^2 = 1 -14/25 = 11/25,
即 (x^4+x^2+1)/x^2 = 11/25,
故 x^2/(x^4+x^2+1) = 25/11.
注意：此题中满足x/(x^2-x+1)=5的x不可能是实数,其实它是一个复数,但是这不影响答题.