Question

已知x/(x2+x+1)=a(a不等于零),求x2/(x4+x2+1)的值

请回答详细

Answer 1

x/(x2+x+1)=a
则1/(x+1/x+1)=a
则x+1/x+1=1/a
则x+1/x=1/a-1
则(x+1/x)²=(1/a-1)²
即x²+1/x²+2=(1/a-1)²
则x²+1/x²+1=(1/a-1)²-1=（1-2a)/a²
若a=1/2，则x²+1/x²+1=0,无解，所以a≠1/2
所以
x2/(x4+x2+1)=1/（x²+1/x²+1)=1/[（1-2a)/a²]=a²/(1-2a)

Answer 2

由已知 (x^2+x+1)/x=1/a
x+1/x+1=1/a
x+1/x=1/a -1
原式的倒数=(x^4+x^2+1)/x^2
=x^2+1+1/x^2
=(x+1/x)^2-1
=(1/a-1)^2-1
=(1-2a)/a^2
所以原式=a^2/(1-2a)