ĐK: \(x\ge\frac{2}{3}\)
\(\sqrt{4x+1}-\sqrt{3x-2}=\frac{x+3}{5}\Leftrightarrow\frac{4x+1-3x+2}{\sqrt{4x+1}+\sqrt{3x-2}}=\frac{x+3}{5}\Leftrightarrow\frac{x+3}{\sqrt{4x+1}+\sqrt{3x-2}}-\frac{x+3}{5}=0\Leftrightarrow\left(x+3\right)\left(\frac{1}{\sqrt{4x+1}+\sqrt{3x-2}}-\frac{1}{5}\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\left(ktm\right)\\\frac{1}{\sqrt{4x+1}+\sqrt{3x-2}}=\frac{1}{5}\end{matrix}\right.\)\(\Leftrightarrow\sqrt{4x+1}+\sqrt{3x-2}=5\Leftrightarrow\sqrt{4x+1}=5-\sqrt{3x-2}\left(\frac{2}{3}\le x\le9\right)\Leftrightarrow4x+1=25+3x-2-10\sqrt{3x-2}\Leftrightarrow22-x=10\sqrt{3x-2}\Leftrightarrow484+x^2-44x=300x-200\Leftrightarrow x^2-344x+684=0\Leftrightarrow\left(x-342\right)\left(x-2\right)=0\Leftrightarrow\)
\(\left[{}\begin{matrix}x=2\left(tm\right)\\x=342\left(ktm\right)\end{matrix}\right.\)
Vậy S={2}
