Đáp án:
\(x = \frac{{13 + \sqrt {97} }}{6}\)
Giải thích các bước giải:
\(\begin{array}{l}
\sqrt {{x^2} + x + 3} = 2x - 3\\
\leftrightarrow \left\{ \begin{array}{l}
2x - 3 \ge 0\\
{x^2} + x + 3 = 4{x^2} - 12x + 9
\end{array} \right.\\
\leftrightarrow \left\{ \begin{array}{l}
x \ge \frac{3}{2}\\
3{x^2} - 13x + 6 = 0
\end{array} \right.\\
\leftrightarrow \left\{ \begin{array}{l}
x \ge \frac{3}{2}\\
\left[ \begin{array}{l}
x = \frac{{13 + \sqrt {97} }}{6}\\
x = \frac{{13 - \sqrt {97} }}{6}
\end{array} \right.
\end{array} \right.\\
\leftrightarrow x = \frac{{13 + \sqrt {97} }}{6}
\end{array}\)
