Đáp án:
\(\left[ \begin{array}{l}
x = 1\\
x = 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\sqrt {\frac{{3x + 1}}{{{x^2} - x + 4}}} + \frac{{3{x^2} + 13}}{{{x^2} - x + 4}} = 5(x \ge \frac{{ - 1}}{3})\\
\Leftrightarrow \frac{{\sqrt {(3x + 1)({x^2} - x + 4)} + 3{x^2} + 13 - 5({x^2} - x + 4)}}{{{x^2} - x + 4}} = 0\\
\Leftrightarrow \sqrt {(3x + 1)({x^2} - x + 4)} - 2{x^2} + 5x - 7 = 0\\
\Leftrightarrow \sqrt {(3x + 1)({x^2} - x + 4)} - 2({x^2} - x + 4) + 3x + 1 = 0\\
\left\{ \begin{array}{l}
\sqrt {{x^2} - x + 4} = a\\
\sqrt {3x + 1} = b
\end{array} \right.\\
\Rightarrow ab - 2{a^2} + {b^2} = 0\\
\Leftrightarrow (a - b)(2a + b) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
a = b\\
2a = - b
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {{x^2} - x + 4} = \sqrt {3x + 1} \\
2\sqrt {{x^2} - x + 4} = - \sqrt {3x + 1}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} - 4x + 3 = 0\\
{x^2} - x + 4 = 3x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 3
\end{array} \right.
\end{array}\)
