Đáp án:

`S={1}` 

Giải thích các bước giải:

`\qquad \sqrt{3x-1}-\sqrt{x+1}=3x^2-2x-1` $(1)$

$ĐK: \begin{cases}3x-1\ge 0\\x+1\ge 0\end{cases}$

`<=>`$\begin{cases}x\ge \dfrac{1}{3}\\x\ge -1\end{cases}$`=>x\ge 1/3`

`(1)<=>{(\sqrt{3x-1}-\sqrt{x+1})(\sqrt{3x-1}+\sqrt{x+1})}/{\sqrt{3x-1}+\sqrt{x+1}}=3x^2-3x+x-1`

`<=>{3x-1-(x-1)}/{\sqrt{3x-1}+\sqrt{x+1}}=3x(x-1)+(x-1)`

`<=>{2(x-1)}/{\sqrt{3x-1}+\sqrt{x+1}}=(x-1)(3x+1)`

`<=>(x-1). [2/{\sqrt{3x-1}+\sqrt{x+1}}-(3x+1)]=0`

`<=>`$\left[\begin{array}{l}x-1=0\\\dfrac{2}{\sqrt{3x-1}+\sqrt{x+1}}-(3x+1)=0\end{array}\right.$

`<=>`$\left[\begin{array}{l}x=1\ (thỏa\ đk)\\\dfrac{2}{\sqrt{3x-1}+\sqrt{x+1}}=3x+1\ (2)\end{array}\right.$

$\\$

Giải `(2)`

Với mọi `x\ge 1/3` ta có:

$\quad \begin{cases}3x-1\ge 0\\x+1\ge \dfrac{4}{3}\end{cases}$

`=>\sqrt{3x-1}+\sqrt{x+1}\ge 0+\sqrt{4/3}=2/{\sqrt{3}}`

`=>2/{\sqrt{3x-1}+\sqrt{x+1}}\le 2 : 2/{\sqrt{3}}=\sqrt{3}<2`

$\\$

`\qquad 3x+1\ge 3. 1/ 3 +1=2`

`=>` Phương trình `(2)` có `VT<2; VP\ge 2`

`=>(2)` vô nghiệm

Vậy phương trình đã cho có tập nghiệm `S={1}`

$\begin{array}{l} ĐKXD:x \ge \dfrac{1}{3}\\ \sqrt {3x - 1}  - \sqrt {x + 1}  = 3{x^2} - 2x - 1\\  \Leftrightarrow 3{x^2} - 2x - 1 + \sqrt {x + 1}  - \sqrt {3x - 1}  = 0\\  \Leftrightarrow \left( {x - 1} \right)\left( {3x + 1} \right) + \dfrac{{x + 1 - 3x + 1}}{{\sqrt {3x - 1}  + \sqrt {x + 1} }} = 0\\  \Leftrightarrow \left( {x - 1} \right)\left( {3x + 1} \right) + \dfrac{{2\left( {1 - x} \right)}}{{\sqrt {3x - 1}  + \sqrt {x + 1} }} = 0\\  \Leftrightarrow \left( {x - 1} \right)\left( {3x + 1 - \dfrac{2}{{\sqrt {3x - 1}  + \sqrt {x + 1} }}} \right) = 0\\  \Leftrightarrow \left[ \begin{array}{l} x = 1\\ 3x + 1 = \dfrac{2}{{\sqrt {3x - 1}  + \sqrt {x + 1} }}\left( * \right) \end{array} \right.\\ \left( * \right) + 3x + 1 \ge 3.\dfrac{1}{3} + 1 = 2\\ \dfrac{2}{{\sqrt {3x - 1}  + \sqrt {x + 1} }} \le \dfrac{2}{{\sqrt {\dfrac{4}{3}} }} = \sqrt 3 \\  \Rightarrow 3x + 1 - \dfrac{2}{{\sqrt {3x - 1}  + \sqrt {x + 1} }} > 0\\ \left( * \right)VN\\  \Rightarrow S = \left\{ 1 \right\}\\ \end{array}$