\(10x^2-9x-8x\sqrt{2x^2-3x+1}+3=0\)
Đặt \(a=\sqrt{2x^2-3x+1}\ge0\) thì:
\(4x^2+3a^2-8ax=0\)
\(\Leftrightarrow\left(2x-a\right)\left(2x-3a\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{a}{2}\\x=\dfrac{3a}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{2x^2-3x+1}}{2}\\x=\dfrac{3\sqrt{2x^2-3x+1}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\sqrt{2x^2-3x+1}\\2x=3\sqrt{2x^2-3x+1}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}4x^2=2x^2-3x+1\\4x^2=9\left(2x^2-3x+1\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x^2+3x-1=0\\\left(3-2x\right)\left(7x-3\right)=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{7}\\x=\dfrac{3}{2}\\x=\dfrac{\sqrt{17}}{4}-\dfrac{3}{4}\end{matrix}\right.\)
