Đáp án:
\[\left[ \begin{array}{l}
x = \frac{{ - 8 + \sqrt {19} }}{{15}}\\
x = \frac{3}{4}
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\[\begin{array}{l}
5\left( {2x + 1} \right)\sqrt {{x^2} + 1} = 10{x^2} + 8x + 4\\
\Leftrightarrow 5\left( {2x + 1} \right)\sqrt {{x^2} + 1} = 2\left( {4{x^2} + 4x + 1} \right) + 2\left( {{x^2} + 1} \right)\\
\Leftrightarrow 2{\left( {2x + 1} \right)^2} - 5\left( {2x + 1} \right)\sqrt {{x^2} + 1} + 2\left( {{x^2} + 1} \right) = 0\\
\Leftrightarrow \left[ {2\left( {2x + 1} \right) - \sqrt {{x^2} + 1} } \right]\left[ {\left( {2x + 1} \right) - 2\sqrt {{x^2} + 1} } \right] = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2\left( {2x + 1} \right) = \sqrt {{x^2} + 1} \\
\left( {2x + 1} \right) = 2\sqrt {{x^2} + 1}
\end{array} \right.\,\,\,\,\,\,\,\,\left( {x \ge - \frac{1}{2}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
4\left( {4{x^2} + 4x + 1} \right) = {x^2} + 1\\
4{x^2} + 4x + 1 = 4\left( {{x^2} + 1} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
15{x^2} + 16x + 3 = 0\\
4x = 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{ - 8 + \sqrt {19} }}{{15}}\left( {t/m} \right)\\
x = \frac{{ - 8 - \sqrt {19} }}{{15}}\left( L \right)\\
x = \frac{3}{4}\left( {t/m} \right)
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \frac{{ - 8 + \sqrt {19} }}{{15}}\\
x = \frac{3}{4}
\end{array} \right.
\end{array}\]
