\(2\sqrt{3x-2}-\sqrt{2x-1}=1\\ \Leftrightarrow2\sqrt{3x-2}=1+\sqrt{2x-1}\\ \Leftrightarrow4\left(3x-2\right)=1+2\sqrt{2x-1}+2x-1\\ \Leftrightarrow12x-8=2\sqrt{2x-1}+2x\\ \Leftrightarrow-2\sqrt{2x-1}=2x-12x+8\\ \Leftrightarrow-2\sqrt{2x-1}=-10x+8\\ \Leftrightarrow\sqrt{2x-1}=5x-4\\ \Leftrightarrow2x-1=25x^2-40x+16\\ \Leftrightarrow2x-1-25x^2+40x-16=0\\ \Leftrightarrow42x-17-25x^2=0\\ \Leftrightarrow-25x^2+42x-17=0\\ \Leftrightarrow25x^2-42x+17=0\\ \Leftrightarrow25x^2-17x-25x+17=0\\ \Leftrightarrow x\left(25x-17\right)-\left(25x-17\right)=0\\ \Leftrightarrow\left(25x-17\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}25x-17=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{25}\\x=1\end{matrix}\right.\)
Thử lại \(x=1\) thỏa mãn
Vậy S = {1}
\(2\sqrt{3x-2}\)-\(\sqrt{2x-1}=1\)
2/3x-2/-/2x-1/=1
2.3x-2-2x+1=1
6x-4-2x+1=1
3x-3=1
3x=4
x=\(\dfrac{4}{3}\)
