Giải pt: `10x^2+3x+1=(6x+1)\sqrt(x^2+3)`

$\begin{array}{l} 10{x^2} + 3x + 1 = \left( {6x + 1} \right)\sqrt {{x^2} + 3} \\  \Leftrightarrow \left( {{x^2} + 3} \right) + \left( {9{x^2} + 3x – 2} \right) – \left( {6x + 1} \right)\sqrt {{x^2} + 3}  = 0 \end{array}$

Đặt $\sqrt{x^2+3}=t(t \ge 0) $. Phương trình trở thành

$\begin{array}{l} {t^2} – \left( {6x + 1} \right)t + 9{x^2} – 3x – 2 = 0\\  \Rightarrow \Delta  = {\left( {6x + 1} \right)^2} – 4\left( {9{x^2} – 3x – 2} \right) = 9\\  \Rightarrow \left[ \begin{array}{l} t = \dfrac{{6x + 1 + \sqrt 9 }}{2} = 3x + 2\\ t = \dfrac{{6x + 1 – \sqrt 9 }}{2} = 3x – 1 \end{array} \right.\\  \Rightarrow \left[ \begin{array}{l} \sqrt {{x^2} + 3}  = 3x + 2\left( {x \ge  – \dfrac{2}{3}} \right)\\ \sqrt {{x^2} + 3}  = 3x – 1\left( {x \ge \dfrac{1}{3}} \right) \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} {x^2} + 3 = 9{x^2} + 12x + 4\\ {x^2} + 3 = 9{x^2} – 6x + 1 \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} 8{x^2} + 12x + 1 = 0\\ 8{x^2} – 6x – 2 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{\sqrt 7  – 3}}{4}\\ x = \dfrac{{ – 3 – \sqrt 7 }}{4}(L)\\ x = 1\\ x =  – \dfrac{1}{4}(L) \end{array} \right.\\  \Rightarrow S = \left\{ {\dfrac{{\sqrt 7  – 3}}{4};1} \right\} \end{array}$