\(\dfrac{x}{x^2-x+1}=\dfrac{1}{5}\)
\(\Rightarrow\dfrac{x^2-x+1}{x}=5\)
\(\Leftrightarrow x-1+\dfrac{1}{x}=5\)
\(\Rightarrow x+\dfrac{1}{x}=6\)
\(\dfrac{x^4+x^2+1}{x^2}=x^2+\dfrac{1}{x^2}+1=\left(x+\dfrac{1}{x}\right)^2-1=6-1=5\)Vậy \(\dfrac{x^2}{x^4+x^2+1}=\dfrac{1}{5}\)
\(\dfrac{x}{x^2-x+1}=\dfrac{1}{5}\Leftrightarrow x^2-x+1=5x\Leftrightarrow x^2-6x+1=0\)
\(\Leftrightarrow\left(x-3\right)^2-8=0\Leftrightarrow\left(x-3\right)^2=8\Leftrightarrow\left\{{}\begin{matrix}x-3=\sqrt{8}\\x-3=-\sqrt{8}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3+\sqrt{8}\\x=3-\sqrt{8}\end{matrix}\right.\) vậy \(x=3+\sqrt{8};x=3-\sqrt{8}\)
th1 : \(x=3+\sqrt{8}\)
thì \(\dfrac{x^2}{x^4+x^2+1}=\dfrac{\left(3+\sqrt{8}\right)^2}{\left(3+\sqrt{8}\right)^4+\left(3+\sqrt{8}\right)^2+1}\)
\(=\dfrac{9+12\sqrt{2}+8}{81+216\sqrt{2}+432+192\sqrt{2}+64+9+12\sqrt{2}+8+1}\)
\(=\dfrac{17+12\sqrt{2}}{595+420\sqrt{2}}=\dfrac{17+12\sqrt{2}}{35\left(17+12\sqrt{2}\right)}=\dfrac{1}{35}\)
th2 : \(x=3-\sqrt{8}\)
thì \(\dfrac{x^2}{x^4+x^2+1}=\dfrac{\left(3-\sqrt{8}\right)^2}{\left(3-\sqrt{8}\right)^4+\left(3-\sqrt{8}\right)^2+1}\)
\(=\dfrac{9-12\sqrt{2}+8}{81-216\sqrt{2}+432-192\sqrt{2}+64+9-12\sqrt{2}+8+1}\)
\(=\dfrac{17-12\sqrt{2}}{595-420\sqrt{2}}=\dfrac{17-12\sqrt{2}}{35\left(17-12\sqrt{2}\right)}=\dfrac{1}{35}\)
vậy \(\dfrac{x^2}{x^4+x^2+1}=\dfrac{1}{35}\) khi \(\dfrac{x}{x^2-x+1}=\dfrac{1}{5}\)
