Ta có :
\(\dfrac{x}{x^2-x+1}=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{x^2-x+1}{x}=\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x^2}{x}-\dfrac{x}{x}+\dfrac{1}{x}=\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x^2}{x}+\dfrac{x}{x}+\dfrac{1}{x}=\dfrac{3}{2}+2\)
\(\Leftrightarrow\dfrac{x^2+x+1}{x}=\dfrac{7}{2}\)
\(\Leftrightarrow\dfrac{x}{x^2+x+1}=\dfrac{2}{7}\)
Lại có :
\(Q=\dfrac{x^2}{x^4+x^2+1}\)
\(=\dfrac{x^2}{x^4+x^3-x^3+x^2+x-x+1}\)
\(=\dfrac{x^2}{x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)}\)
\(=\dfrac{x^2}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x}{x^2+x+1}.\dfrac{x}{x^2-x+1}\)
\(=\dfrac{2}{3}.\dfrac{2}{7}\)
\(=\dfrac{4}{21}\)
Vậy...
\(\dfrac{x}{x^2-x+1}=\dfrac{2}{3}\Leftrightarrow\dfrac{x^2-x+1}{x}=\dfrac{3}{2}\)
\(\Leftrightarrow x-1+\dfrac{1}{x}=\dfrac{3}{2}\)
\(\Leftrightarrow x+\dfrac{1}{x}=\dfrac{5}{2}\)
\(\Leftrightarrow x^2+\dfrac{1}{x^2}+2=\dfrac{25}{4}\Leftrightarrow x^2+\dfrac{1}{x^2}=\dfrac{17}{4}\)
\(\Leftrightarrow x^2+\dfrac{1}{x^2}+1=\dfrac{21}{4}\)
\(\dfrac{1}{Q}=\dfrac{x^4+x^2+1}{x^2}=x^2+1+\dfrac{1}{x^2}=\dfrac{17}{4}\Leftrightarrow Q=\dfrac{4}{17}\)
