若x/x^2-3x+1=2,求分式x^2/x^4+x^2+1的值本题用常规法无法解答,要用倒数带入.

x/(x^2-3x+1)=2
2(x^2-3x+1)=x
2x^2-6x+2=x
2x^2-7x+2=0
同时除以x，得：
2x-7+2/x=0
2(x+1/x)=7
x+1/x=7/2
平方，得：
x^2+2+1/x^2=49/4
x^2+1/x^2=41/4
x^2/x^4+x^2+1
=1/(x^2+1+1/x^2)
=1/(41/4+1)
=1/(45/4)
=4/45

x/x^2-3x+1=2
则(x^2-3x+1)/x=1/2
所以x-3+1/x=1/2
所以x+1/x=7/2
两边平方,得x²+1/x²+2=49/4
所以x²+1/x²=41/4
分式x^2/x^4+x^2+1=1/(x²+1+1/x²)=1/(41/4+1)=4/45