Đáp án:
\(\left[ \begin{array}{l}
x = 1\\
x = - 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:5{x^2} + 10x + 1 \ge 0 \to \left[ \begin{array}{l}
x \ge \dfrac{{ - 5 + 2\sqrt 5 }}{5}\\
x \le \dfrac{{ - 5 - 2\sqrt 5 }}{5}
\end{array} \right.\\
\sqrt {5{x^2} + 10x + 1} = 7 - \left( {{x^2} + 2x} \right)\\
Đặt:\sqrt {5{x^2} + 10x + 1} = t\left( {t \ge 0} \right)\\
\to 5{x^2} + 10x + 1 = {t^2}\\
\to 5\left( {{x^2} + 2x} \right) = {t^2} - 1\\
\to {x^2} + 2x = \dfrac{{{t^2} - 1}}{5}\\
Pt \to t = 7 - \dfrac{{{t^2} - 1}}{5}\\
\to 5t = 35 - {t^2} + 1\\
\to {t^2} + 5t - 36 = 0\\
\to \left[ \begin{array}{l}
t = 4\\
t = - 9\left( l \right)
\end{array} \right.\\
\to 5{x^2} + 10x + 1 = 16\\
\to 5{x^2} + 10x - 15 = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 3
\end{array} \right.\left( {TM} \right)
\end{array}\)
