Đặt \(t=\sqrt{3x^2+5x+1}\left(t\ge0\right)\)
pt đã cho trở thành: \(\sqrt{t^2+7}-t=1\Leftrightarrow\sqrt{t^2+7}=t+1\)
- bình phương 2 vế, giải ra t, trả lại nghiệm x, tìm x
\(\sqrt {3{x^2} + 5x + 8} - \sqrt {3{x^2} + 5x + 1} = 1\\ \text{Điều kiện}: \forall x \in \mathbb{R}\\ \text{Đặt}:\sqrt {3{x^2} + 5x + 8} =a; \sqrt {3{x^2} + 5x + 1} = b\\ \Rightarrow \left\{ \begin{array}{l} a - b = 1\\ {a^2} - {b^2} = 7 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} a = b + 1\\ {\left( {b + 1} \right)^2} - {b^2} = 7 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} a = b + 1\\ 2b = 6 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} a = 4\\ b = 3 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} \sqrt {3{x^2} + 5x + 8} = 4\\ \sqrt {3{x^2} + 5x + 1} = 3 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 3{x^2} + 5x + 8 = 16\\ 3{x^2} + 5x + 1 = 9 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 3{x^2} + 5x - 8 = 0\\ 3{x^2} + 5x - 8 = 0 \end{array} \right.\\ \Leftrightarrow \left( {x - 1} \right)\left( {3x + 8} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 1\\ x = - \dfrac{8}{3} \end{array} \right. \)
\(\text{Cách khác:}\\ \text{Đặt}:t = \sqrt {3{x^2} + 5x + 8} \left( {t \ge 0} \right) \text{thì} t^2=3x^2+5x+8\\ PT:t-\sqrt{t^2-7}=1 \Leftrightarrow \sqrt{t^2-7} = t- 1 \)
\( \Leftrightarrow \left\{ \begin{array}{l} t - 1 \ge 0\\ {t^2} - 7 = {\left( {t - 1} \right)^2} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} t \ge 1\\ 2t = 8 \end{array} \right. \Leftrightarrow t = 4\)
\(\text{Do đó:}\) \(\sqrt{3x^2+5x+8}=4\Leftrightarrow3x^2+5x-8=0\Leftrightarrow\left(x-1\right)\left(3x+8\right)=0\Leftrightarrow\)\(x=1 \text{hoặc} x =- \dfrac{8}{3}\)