Đáp án:
\(\left[ \begin{array}{l}
x \ge 34\\
x \le 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:\left[ \begin{array}{l}
x \ge 32\\
x \le 2
\end{array} \right.\\
6\sqrt {\left( {x - 2} \right)\left( {x - 32} \right)} \le {x^2} - 34x + 48\\
\to 6\sqrt {{x^2} - 34x + 64} \le {x^2} - 34x + 48\\
Đặt:\sqrt {{x^2} - 34x + 64} = t\left( {t \ge 0} \right)\\
\to {x^2} - 34x + 64 = {t^2}\\
\to {x^2} - 34x = {t^2} - 64\\
Bpt \to 6t \le {t^2} - 64 + 48\\
\to {t^2} - 6t - 16 \ge 0\\
\to \left[ \begin{array}{l}
t \ge 8\\
t \le - 2\left( l \right)
\end{array} \right.\\
\to \sqrt {{x^2} - 34x + 64} \ge 8\\
\to {x^2} - 34x + 64 \ge 64\\
\to {x^2} - 34x \ge 0\\
\to x\left( {x - 34} \right) \ge 0\\
\to \left[ \begin{array}{l}
x \ge 34\\
x \le 0
\end{array} \right.\\
KL:\left[ \begin{array}{l}
x \ge 34\\
x \le 0
\end{array} \right.
\end{array}\)
