已知x/x2-x+1=1/7,求x2/x4+x2+1
问题描述：
已知x/x2-x+1=1/7,求x2/x4+x2+1
答
x/(x^2-x+1)=1/7(取倒数)
(x^2-x+1)/x=7
x-1+1/x=7
x+1/x=8
x^2/(x^4+x^2+1)(取倒数)
(x^4+x^2+1)/x^2
=x^2+1+1/x^2
=(x+1/x)^2-1
=8^2-1
=63
(取倒数)
所以：x^2/(x^4+x^2+1)=1/63
答
x/(x²-x+1)=1/7
x²-x+1=7x
x²+1=8x
两边平方
x4+2x²+1=64x²
x4+1=62x²
所以原式=x²/(62²+x²)=1/63