Đáp án:
\(3 \le x \le \dfrac{{19}}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:\left[ \begin{array}{l}
x \ge 3\\
x \le 2
\end{array} \right.\\
\left( {2x - 5 - \sqrt {{x^2} - x + 25} } \right)\left( {\sqrt {{x^2} - 5x + 6} } \right) \le 0\\
\to 2x - 5 - \sqrt {{x^2} - x + 25} \le 0Do:\left( {\sqrt {{x^2} - 5x + 6} \ge 0\forall \left[ \begin{array}{l}
x \ge 3\\
x \le 2
\end{array} \right.} \right)\\
\to 2x - 5 \le \sqrt {{x^2} - x + 25} \\
\to 4{x^2} - 20x + 25 \le {x^2} - x + 25\left( {DK:x \ge \dfrac{5}{2}} \right)\\
\to 3{x^2} - 19x \le 0\\
\to 0 \le x \le \dfrac{{19}}{3}\\
KL:3 \le x \le \dfrac{{19}}{3}
\end{array}\)
