ĐKXĐ : \(x\ge5\)
Ta có \(x-3\sqrt{x}+4=2\sqrt{x-5}\)
\(\Leftrightarrow x-3\sqrt{x}=2\left(\sqrt{x-5}-2\right)\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-3\right)=2.\dfrac{x-9}{\sqrt{x-5}+2}\)
\(\Leftrightarrow\sqrt{x}.\left(\sqrt{x}-3\right)=\dfrac{2\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}{\sqrt{x-5}+2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3=0\\\sqrt{x}=\dfrac{2.\left(\sqrt{x}+3\right)}{\sqrt{x-5}+2}\end{matrix}\right.\)
Với \(\sqrt{x}-3=0\Leftrightarrow x=9\left(tm\right)\)
Với \(\sqrt{x}=\dfrac{2.\left(\sqrt{x}+3\right)}{\sqrt{x-5}+2}\Leftrightarrow\sqrt{x}.\sqrt{x-5}=6\)
\(\Leftrightarrow x^2-5x-36=0\Leftrightarrow\left[{}\begin{matrix}x=9\left(tm\right)\\x=-4\left(\text{loại}\right)\end{matrix}\right.\)
Tập nghiệm \(S=\left\{9\right\}\)
