已知x/x²-x+1=7,求x²/x的四次方+x的平方+1 的解

网友回答

x/(x^2-x+1)=7
==>(x^2-x+1)/x=1/7
==>x-1+1/x=1/7
==>x+1/x=8/7
x^2/(x^4+x^2+1)
=x^2/[(x^2+1)^2-x^2]
=x^2/(x^2+x+1)(x^2-x+1)
=7x/(x^2+x+1)
=7/(x+1+1/x)
=7/(15/7)
=49/15.
======以下答案可供参考======
供参考答案1:
∵x/(x²-x+1)=7
∴(x²-x+1)/x=1/7
∴(x-1+1/x)=1/7
∴x+1/x=1/7+1=8/7
∵x^2/(x^4+x^2+1)
=1/(x²+1+1/x²)
=1/[(x²+1/x²+2)-1]
=1/[(x+1/x)²-1]
=1/[(8/7)²-1]
=1/(64/49-1)
=1/(15/49)
=49/15