Đáp án:
\[\left[ \begin{array}{l}
x = 3\\
x = 8
\end{array} \right.\]
Giải thích các bước giải:
ĐKXĐ: \(x \ge \frac{8}{3}\)
Ta có:
\(\begin{array}{l}
\sqrt {3x - 8} - \sqrt {x + 1} = \frac{{2x - 11}}{5}\\
\Leftrightarrow \left( {\sqrt {3x - 8} - \frac{{3x - 4}}{5}} \right) + \left( {\frac{{x + 7}}{5} - \sqrt {x + 1} } \right) = 0\\
\Leftrightarrow \left( {5\sqrt {3x - 8} - \left( {3x - 4} \right)} \right) + \left( {\left( {x + 7} \right) - 5\sqrt {x + 1} } \right) = 0\\
\Leftrightarrow \frac{{25\left( {3x - 8} \right) - \left( {9{x^2} - 24x + 16} \right)}}{{5\sqrt {3x - 8} + 3x - 4}} + \frac{{{x^2} + 14x + 49 - 25\left( {x + 1} \right)}}{{x + 7 + 5\sqrt {x + 1} }} = 0\\
\Leftrightarrow \frac{{ - 9{x^2} + 99x - 216}}{{5\sqrt {3x - 8} + 3x - 4}} + \frac{{{x^2} - 11x + 24}}{{x + 7 + 5\sqrt {x + 1} }} = 0\\
\Leftrightarrow \left( {{x^2} - 11x + 24} \right)\left( {\frac{{ - 9}}{{5\sqrt {3x - 8} + 3x - 4}} + \frac{1}{{x + 7 + 5\sqrt {x + 1} }}} \right) = 0\\
\Leftrightarrow {x^2} - 11x + 24 = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {x - 8} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 8
\end{array} \right.
\end{array}\)
