ĐK: \(\left[{}\begin{matrix}x=0,3,4\\x>4\end{matrix}\right.\)
pt \(\Leftrightarrow\sqrt{x\left(x-3\right)}+\sqrt{x\left(x-4\right)}=2x\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x-3}-\sqrt{x}\right)=\sqrt{x}\left(\sqrt{x}-\sqrt{x-4}\right)\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x-3}-\sqrt{x}-\sqrt{x}+\sqrt{x-4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x-3}-\sqrt{x}=\sqrt{x}-\sqrt{x-4}\left(1\right)\end{matrix}\right.\)
Vì \(x\ge4\Rightarrow x>x-3\Rightarrow\sqrt{x-3}< \sqrt{x}\Rightarrow\sqrt{x-3}-\sqrt{x}< 0\)
mà \(x\ge4\Rightarrow x>x-4\Rightarrow\sqrt{x}>\sqrt{x-4}\Rightarrow\sqrt{x}-\sqrt{x-4}>0\)
Vậy phương trình (1) vô nghiệm
KL: x = 0
