\(\frac{x}{\sqrt{3x-2}}+\frac{\sqrt{3x-2}}{x}=2\)
Đk:\(\sqrt{3x-2}\ge0\Rightarrow3x-2\ge0\Rightarrow x\ge\frac{2}{3}\)
\(\Leftrightarrow\frac{x^2}{x\sqrt{3x-2}}+\frac{3x-2}{x\sqrt{3x-2}}-\frac{2\left(x\sqrt{3x-2}\right)}{x\sqrt{3x-2}}=0\)
\(\Leftrightarrow\frac{x^2+3x-2-2\left(x\sqrt{3x-2}\right)}{x\sqrt{3x-2}}=0\)
\(\Leftrightarrow x^2+3x-2-2\left(x\sqrt{3x-2}\right)=0\)
\(\Leftrightarrow x^2+3x-2=2x\sqrt{3x-2}\)
\(\Leftrightarrow\left(x^2+3x-2\right)^2=\left(2x\right)^2\sqrt{\left(3x-2\right)^2}\)
\(\Leftrightarrow x^4+6x^3+5x^2-12x+4=4x^2\left(3x-2\right)\)
\(\Leftrightarrow x^4+6x^3+5x^2-12x+4=12x^3-8x^2\)
\(\Leftrightarrow x^4-6x^3+13x^2-12x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2\left(x-1\right)^2=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(x-2\right)^2=0\\\left(x-1\right)^2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x-1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=1\end{array}\right.\)(thỏa mãn)
Vậy pt có nghiệm là \(\left[\begin{array}{nghiempt}x=2\\x=1\end{array}\right.\)