ĐKXĐ: \(x\ge1\)
\(\sqrt{5x-1}=\sqrt{3x-2}+\sqrt{x-1}\)
\(\Leftrightarrow5x-1=3x-2+2\sqrt{\left(3x-2\right)\left(x-1\right)}+x-1\)
\(\Leftrightarrow x+2=2\sqrt{3x^2-5x+2}\)
\(\Leftrightarrow x^2+4x+4=4\left(3x^2-5x+2\right)\)
\(\Leftrightarrow11x^2-24x+4=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{2}{11}\end{matrix}\right.\)
ĐKXĐ: \((x\ge1)\)
\(\sqrt{5x-1}=\sqrt{3x-2}+\sqrt{x-1}\)
\(\Leftrightarrow5x-1=3x-2+2\sqrt{\left(3x-2\right)\left(x-1\right)}+x-1\)
\(\Leftrightarrow x+2=2\sqrt{\left(3x-2\right)\left(x-1\right)}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2\ge0\\\left(x+2\right)^2=2\sqrt{\left(3x-2\right)\left(x-1\right)}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^2+4x+4=4\left(3x-2\right)\left(x-1\right)\\ \Leftrightarrow x^2+4x+4=12x^2-20x+8\\ \Leftrightarrow x^2+4x+4-12x^2+20x-8=0\\ \Leftrightarrow-11x^2+24x-4=0\\ \Leftrightarrow11x^2-24x+4=0\\ \Leftrightarrow11x^2-2x-22x+4=0\\ \Leftrightarrow x\left(11x-2\right)-2\left(11x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(11x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\11x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=\dfrac{2}{11}\left(KTM\right)\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)
