Đáp án:
\[\left[ \begin{array}{l}
x = 3\\
x = 7
\end{array} \right.\]
Giải thích các bước giải:
ĐKXĐ:\(x \ge \frac{5}{2}\)
Ta có:
\(\begin{array}{l}
2\sqrt {2x - 5} + 2\sqrt {3x - 5} = {x^2} - 8x + 21\\
\Leftrightarrow \left( {2\sqrt {2x - 5} - \left( {x - 1} \right)} \right) + \left( {2\sqrt {3x - 5} - \left( {x + 1} \right)} \right) = {x^2} - 8x + 21 - \left( {x - 1} \right) - \left( {x + 1} \right)\\
\Leftrightarrow \frac{{4\left( {2x - 5} \right) - {{\left( {x - 1} \right)}^2}}}{{2\sqrt {2x - 5} + x - 1}} + \frac{{4\left( {3x - 5} \right) - {{\left( {x + 1} \right)}^2}}}{{2\sqrt {3x - 5} + x + 1}} = {x^2} - 10x + 21\\
\Leftrightarrow \frac{{ - {x^2} + 10x - 21}}{{2\sqrt {2x - 5} + x - 1}} + \frac{{ - {x^2} + 10x - 21}}{{2\sqrt {3x - 5} + x + 1}} = {x^2} - 10x + 21\\
\Leftrightarrow \left( {{x^2} - 10x + 21} \right)\left( {1 + \frac{1}{{2\sqrt {2x - 5} + x - 1}} + \frac{1}{{2\sqrt {3x - 5} + x + 1}}} \right) = 0\\
x \ge \frac{5}{2} \Rightarrow 1 + \frac{1}{{2\sqrt {2x - 5} + x - 1}} + \frac{1}{{2\sqrt {3x - 5} + x + 1}} > 0\\
\Rightarrow {x^2} - 10x + 21 = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {x - 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 7
\end{array} \right.\left( {t/m} \right)
\end{array}\)
