Đáp án:
x=2
Giải thích các bước giải:
$\begin{array}{l}
\sqrt {x + 2} + \sqrt {4x + 1} = 5\\
DK:\left\{ \begin{array}{l}
x + 2 \ge 0\\
4x + 1 \ge 0
\end{array} \right. \Leftrightarrow x \ge - \dfrac{1}{4}\\
PT \Leftrightarrow \sqrt {x + 2} = 5 - \sqrt {4x + 1} \\
\Leftrightarrow \left\{ \begin{array}{l}
5 - \sqrt {4x + 1} \ge 0\\
x + 2 = 25 + 4x + 1 - 10\sqrt {4x + 1}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt {4x + 1} \le 5\\
10\sqrt {4x + 1} = 3x + 24
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
4x + 1 \le 25\\
100\left( {4x + 1} \right) = 9{x^2} + 144x + 576
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le 6\\
9{x^2} - 256x + 476 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le 6\\
\left[ \begin{array}{l}
x = \dfrac{{238}}{9}\\
x = 2
\end{array} \right.
\end{array} \right. \Leftrightarrow x = 2
\end{array}$
