1. a) Ta có: \(x^2-2y^2=xy\) \(\Leftrightarrow\) \(x^2-xy-2y^2=0\)

\(\Leftrightarrow\) \(x^2+xy-2xy-2y^2=0\)

\(\Leftrightarrow\) \(x\left(x+y\right)-2y\left(x+y\right)=0\)

\(\Leftrightarrow\) \(\left(x+y\right)\left(x-2y\right)=0\)

Vì \(\left(x+y\right)\ne0\) nên \(x-2y=0\) hay \(x=2y\). Thay \(x=2y\) vào A, ta được:

\(A=\dfrac{\left(2y\right)^2-y^2}{\left(2y\right)^2+y^2}=\dfrac{4y^2-y^2}{4y^2+y^2}=\dfrac{3y^2}{5y^2}=\dfrac{3}{5}\)

Từ \(\dfrac{x}{x^2-x+1}=\dfrac{1}{5}\Rightarrow x^2-x+1=5x\Rightarrow x^2+1=5x+x=6x\)

Ta có: \(E=\dfrac{x^2}{x^4+x^2+1}\)

= \(\dfrac{x^2}{x^4-x+x^2+x+1}\)

= \(\dfrac{x^2}{x\left(x^3-1\right)+\left(x^2+x+1\right)}\)

= \(\dfrac{x^2}{x\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)}\)

= \(\dfrac{x^2}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\)

= \(\dfrac{x}{x^2+x+1}.\dfrac{x}{x^2-x+1}\)

= \(\dfrac{x}{6x+x}.\dfrac{1}{5}\) = \(\dfrac{x}{5.7x}\) = \(\dfrac{1}{35}\)