\(\dfrac{x}{x^2-x+1}=\dfrac{2}{3}\)
\(\Leftrightarrow3x=2x^2-2x+2\)
\(\Leftrightarrow2x^2-5x+2=0\)
\(\Leftrightarrow2x^2-4x-x+2=0\)
\(\Leftrightarrow2x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)
Trường hợp x=2
\(Q=\dfrac{x^2}{x^4+x^2+1}=\dfrac{2^2}{2^4+2^2+1}=\dfrac{4}{21}\)
Trường hợp \(x=\dfrac{1}{2}\)
\(Q=\dfrac{x^2}{x^4+x^2+1}=\dfrac{\left(\dfrac{1}{2}\right)^2}{\left(\dfrac{1}{2}\right)^4+\left(\dfrac{1}{2}\right)^2+1}=\dfrac{\dfrac{1}{4}}{\dfrac{21}{16}}=\dfrac{4}{21}\)
