ĐK: x>0
\(\dfrac{x-1}{\sqrt{x}}=\dfrac{3}{2}\Leftrightarrow2\left(x-1\right)=3\sqrt{x}\Leftrightarrow2x-2=3\sqrt{x}\Leftrightarrow2x-3\sqrt{x}-2=0\Leftrightarrow2x+\sqrt{x}-4\sqrt{x}-2=0\Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)=0\Leftrightarrow\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=0\\2\sqrt{x}+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\2\sqrt{x}=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\\sqrt{x}=-\dfrac{1}{2}\left(ktm\right)\end{matrix}\right.\)
Vậy x=4 là nghiệm của phương trình
