Đáp án:
\[x = 5\]
Giải thích các bước giải:
\(\begin{array}{l}
\left( { - \frac{1}{3} \le x \le 6} \right)\\
\sqrt {3x + 1} - \sqrt {6 - x} + 3{x^2} - 14x - 8 = 0\\
\Leftrightarrow \left( {\sqrt {3x + 1} - 4} \right) + \left( {1 - \sqrt {6 - x} } \right) + 3{x^2} - 14x - 5 = 0\\
\Leftrightarrow \frac{{3x + 1 - 16}}{{\sqrt {3x + 1} + 4}} + \frac{{1 - 6 + x}}{{1 + \sqrt {6 - x} }} + \left( {x - 5} \right)\left( {3x + 1} \right) = 0\\
\Leftrightarrow \frac{{3\left( {x - 5} \right)}}{{\sqrt {3x + 1} + 4}} + \frac{{x - 5}}{{\sqrt {6 - x} + 1}} + \left( {x - 5} \right)\left( {3x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
\frac{3}{{\sqrt {3x + 1} + 4}} + \frac{1}{{\sqrt {6 - x} + 1}} + 3x + 1 = 0
\end{array} \right.\\
- \frac{1}{3} \le x \le 6 \Rightarrow \frac{3}{{\sqrt {3x + 1} + 4}} + \frac{1}{{\sqrt {6 - x} + 1}} + 3x + 1 > 0\\
\Rightarrow x = 5
\end{array}\)
