ĐKXĐ : \(\left\{{}\begin{matrix}\frac{x-1}{x}\ge0\\x\ne0\end{matrix}\right.\) => \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1\ge0\\x>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1\le0\\x< 0\end{matrix}\right.\end{matrix}\right.\) => \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge1\\x>0\end{matrix}\right.\\\left[{}\begin{matrix}x\le1\\x< 0\end{matrix}\right.\end{matrix}\right.\) => \(\left[{}\begin{matrix}x\ge1\\x< 0\end{matrix}\right.\)
Với \(x\ge1\)
=> \(x\sqrt{\frac{x-1}{x}}\ge0\)
=> \(x-2\ge0\)
=> \(x\ge2\)
=> \(\left[{}\begin{matrix}x\ge2\\x< 0\end{matrix}\right.\)
Ta có : \(x\sqrt{\frac{x-1}{x}}=x-2\)
=> \(x^2\left(\frac{x-1}{x}\right)=x^2-4x+4\)
=> \(x\left(x-1\right)=x^2-4x+4\)
=> \(x^2-4x+4=x^2-x\)
=> \(3x=4\)
=> \(x=\frac{4}{3}\left(L\right)\)
Vậy phương trình vô nghiệm .
