ĐKXĐ: \(x>\dfrac{2}{3}\)
\(\dfrac{x^2}{\sqrt{3x-2}}-\sqrt{3x-2}=1-x\)
\(\Leftrightarrow\dfrac{x^2-3x+2}{\sqrt{3x-2}}+x-1=0\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{\sqrt{3x-2}}+x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{x-2}{\sqrt{3x-2}}+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\Rightarrow x=1\\\dfrac{x-2}{\sqrt{3x-2}}+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x-2+\sqrt{3x-2}=0\Leftrightarrow\sqrt{3x-2}=2-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}2-x\ge0\\3x-2=\left(2-x\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le2\\x^2-7x+6=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\x=6>2\left(l\right)\end{matrix}\right.\)
Vậy pt đã cho có nghiệm duy nhất \(x=1\)
