Đáp án: x=1 hoặc x=-1
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:\left\{ \begin{array}{l}
2{x^2} + 8x + 6 \ge 0\\
{x^2} - 1 \ge 0\\
2x + 2 \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {x + 1} \right)\left( {2x + 6} \right) \ge 0\\
\left( {x + 1} \right)\left( {x - 1} \right) \ge 0\\
2\left( {x + 1} \right) \ge 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = - 1\\
x \ge 1
\end{array} \right.\\
\sqrt {2{x^2} + 8x + 6} + \sqrt {{x^2} - 1} = 2x + 2\\
\Rightarrow \sqrt {\left( {x + 1} \right).\left( {2x + 6} \right)} + \sqrt {\left( {x + 1} \right)\left( {x - 1} \right)} - 2\left( {x + 1} \right) = 0\\
\Rightarrow \sqrt {x + 1} .\left( {\sqrt {2x + 6} + \sqrt {x - 1} - 2\sqrt {x + 1} } \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt {x + 1} = 0\\
\sqrt {2x + 6} + \sqrt {x - 1} - 2\sqrt {x + 1} = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x + 1 = 0\\
\sqrt {2x + 6} + \sqrt {x - 1} = 2\sqrt {x + 1}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - 1\\
2x + 6 + x - 1 + 2\sqrt {\left( {2x + 6} \right)\left( {x - 1} \right)} = 4\left( {x + 1} \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - 1\\
x - 1 = 2\sqrt {2{x^2} + 4x - 6}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - 1\\
{x^2} - 2x + 1 = 4\left( {2{x^2} + 4x - 6} \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - 1\left( {ktm} \right)\\
7{x^2} + 18x - 25 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = - \frac{{25}}{7}\left( {ktm} \right)
\end{array} \right.\\
Vay\,x = 1\,hoac\,x = - 1
\end{array}$
