a) ĐKXĐ: \(x\ge1\)
Ta có: \(\sqrt{5x-5}-\sqrt{35}=0\)
\(\Leftrightarrow\sqrt{5x-5}=\sqrt{35}\)
\(\Leftrightarrow\left(\sqrt{5x-5}\right)^2=\left(\sqrt{35}\right)^2\)
\(\Leftrightarrow\left|5x-5\right|=35\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-5=35\\5x-5=-35\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=40\\5x=-30\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-6\left(loại\right)\end{matrix}\right.\)
Vậy: S={8}
b) ĐKXĐ: \(x\ge3\)
Ta có: \(\sqrt{x^2-9}=3\sqrt{x-3}\)
\(\Leftrightarrow\sqrt{x-3}\cdot\sqrt{x+3}-3\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}\cdot\left(\sqrt{x+3}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\x=6\left(nhận\right)\end{matrix}\right.\)
Vậy: S={3;6}
