Đáp án:
$S=\{2\}$
Giải thích các bước giải:
ĐKXĐ: $\begin{cases}5x-1\ge 0\\3x-2\ge 0\\x-1\ge 0\end{cases}⇒\begin{cases}x\ge \dfrac{1}{5}\\x\ge \dfrac{2}{3}\\x\ge 1\end{cases}⇒x\ge 1$
$\sqrt{5x-1}-\sqrt{3x-2}-\sqrt{x-1}=0$
$⇔\sqrt{5x-1}=\sqrt{3x-2}+\sqrt{x-1}$
$⇔5x-1=3x-2+x-1+2\sqrt{(3x-2)(x-1)}$
$⇔x+2=2\sqrt{3x^2-5x+2}$
$⇔(x+2)^2=4(3x^2-5x+2)$
$⇔x^2+4x+4=12x^2-20x+8$
$⇔11x^2-24x+4=0$
$⇔11x^2-22x-2x+4=0$
$⇔(x-2)(11x-2)=0$
$⇔\left[ \begin{array}{l}x-2=0\\11x-2=0\end{array} \right.⇔\left[ \begin{array}{l}x=2\\x=\dfrac{2}{11}\,(L)\end{array} \right.$
Vậy $S=\{2\}$.
