ĐK: \(x\ge\frac{1}{5}\)
\(\sqrt{5x-1}-\sqrt{3x+13}=\frac{x-7}{3}\Leftrightarrow\frac{\left(\sqrt{5x-1}-\sqrt{3x+13}\right)\left(\sqrt{5x-1}+\sqrt{3x+13}\right)}{\sqrt{5x-1}+\sqrt{3x+13}}=\frac{x-7}{3}\Leftrightarrow\frac{5x-1-3x-13}{\sqrt{5x-1}+\sqrt{3x+13}}=\frac{x-7}{3}\Leftrightarrow\frac{2x-14}{\sqrt{5x-1}+\sqrt{3x+13}}=\frac{x-7}{3}\Leftrightarrow\frac{2\left(x-7\right)}{\sqrt{5x-1}+\sqrt{3x+13}}-\frac{x-7}{3}=0\Leftrightarrow\left(x-7\right)\left(\frac{2}{\sqrt{5x-1}+\sqrt{3x+13}}-\frac{1}{3}\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-7=0\\\frac{2}{\sqrt{5x-1}+\sqrt{3x+13}}-\frac{1}{3}=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=7\left(tm\right)\\\frac{2}{\sqrt{5x-1}+\sqrt{3x+13}}=\frac{1}{3}\left(1\right)\end{matrix}\right.\)
(1)\(\Leftrightarrow6=\sqrt{5x-1}+\sqrt{3x+13}\Leftrightarrow\sqrt{5x-1}=6-\sqrt{3x+13}\Leftrightarrow5x-1=36-12\sqrt{3x+13}+3x+13\Leftrightarrow50-2x=12\sqrt{3x+13}\left(x\le25\right)\Leftrightarrow25-x=6\sqrt{3x+13}\Leftrightarrow625-50x+x^2=108x+468\Leftrightarrow x^2-158x+157=0\Leftrightarrow\left(x-157\right)\left(x-1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-157=0\\x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=157\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
Vậy S={1;7}
