Đáp án:
x=-2
Giải thích các bước giải:
$\begin{array}{l}
3{x^2} + 4x + 10 = 2\sqrt {14{x^2} - 7} \\
(dkxd:14{x^2} - 7 \ge 0\\
\Leftrightarrow {x^2} \ge \frac{7}{{14}} = \frac{1}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge \frac{{\sqrt 2 }}{2}\\
x \le - \frac{{\sqrt 2 }}{2}
\end{array} \right.\\
pt \Leftrightarrow 3{x^2} + 4x + 10 = 2.\sqrt 7 .\sqrt {2{x^2} - 1} \\
\Leftrightarrow 2{x^2} - 1 - 2.\sqrt 7 .\sqrt {2{x^2} - 1} + 7 + {x^2} + 4x + 4 = 0\\
\Leftrightarrow {\left( {\sqrt {2{x^2} - 1} - \sqrt 7 } \right)^2} + {\left( {x + 2} \right)^2} = 0\\
\Rightarrow \left\{ \begin{array}{l}
\sqrt {2{x^2} - 1} - \sqrt 7 = 0\\
x + 2 = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
2{x^2} - 1 = 7\\
x = - 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} = 4\\
x = - 2
\end{array} \right. \Rightarrow x = - 2(tmdk)
\end{array}$
