Đáp án:
\(x=1.\)
Giải thích các bước giải:
\[\begin{array}{l}
\,\,\,\,\,\,\,4{x^2} + 3x + 3 = 4x\sqrt {x + 3} + 2\sqrt {2x - 1} \,\,\,\,\left( {DK:\,\,\,x \ge \frac{1}{2}} \right)\\
\Leftrightarrow 4{x^2} - 4x\sqrt {x + 3} + x + 3 + 2x - 1 - 2\sqrt {2x - 1} + 1 = 0\\
\Leftrightarrow {\left( {2x - \sqrt {x + 3} } \right)^2} + {\left( {\sqrt {2x - 1} - 1} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
2x - \sqrt {x + 3} = 0\\
\sqrt {2x - 1} - 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x = \sqrt {x + 3} \\
\sqrt {2x - 1} = 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
4{x^2} = x + 3\\
2x - 1 = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
4{x^2} - x - 3 = 0\\
2x = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x = 1\\
x = - \frac{3}{4}\,\,\,\left( {ktm} \right)
\end{array} \right.\\
x = 1
\end{array} \right. \Leftrightarrow x = 1\,\,\left( {tm} \right).
\end{array}\]
