Ta biến đổi biểu thức hai vế trước cho gọn
\(\sqrt{x^4+x^2+1}=\sqrt{x^4+2x^2+1-x^2}=\sqrt{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)
\(x^2-3x+1=2\left(x^2-x+1\right)-\left(x^2+x+1\right)\)
OK, đặt \(\sqrt{x^2-x+1}=a>0\) ; \(\sqrt{x^2+x+1}=b>0\) , pt tương đương:
\(2a^2-b^2=\dfrac{-\sqrt{3}}{3}ab\Leftrightarrow2\dfrac{a}{b}-\dfrac{b}{a}+\dfrac{\sqrt{3}}{3}=0\)
Đặt \(\dfrac{a}{b}=t>0\Rightarrow2t-\dfrac{1}{t}+\dfrac{\sqrt{3}}{3}=0\Leftrightarrow2t^2+\dfrac{\sqrt{3}}{3}t-1=0\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{\sqrt{3}}{3}\\t=\dfrac{-\sqrt{3}}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{\dfrac{x^2-x+1}{x^2+x+1}}=\dfrac{\sqrt{3}}{3}\Leftrightarrow\dfrac{x^2-x+1}{x^2+x+1}=\dfrac{1}{3}\)
\(\Leftrightarrow2x^2-4x+2=0\Leftrightarrow2\left(x-1\right)^2=0\Leftrightarrow x=1\)
