Đáp án:
$\begin{array}{l}
c)Dkxd:3 - x \ge 0 \Leftrightarrow x \le 3\\
\sqrt {{{\left( {x - 3} \right)}^2}} = 3 - x\\
\Leftrightarrow \left| {x - 3} \right| = 3 - x\\
\Leftrightarrow x - 3 \le 0\\
\Leftrightarrow x \le 3\\
Vậy\,x \le 3\\
d)\sqrt {4{x^2} - 20x + 25} + 2x = 5\\
\Leftrightarrow \sqrt {{{\left( {2x - 5} \right)}^2}} = 5 - 2x\\
Dkxd:5 - 2x \ge 0 \Leftrightarrow x \le \dfrac{5}{2}\\
\Leftrightarrow \left| {2x - 5} \right| = 5 - 2x\\
\Leftrightarrow 2x - 5 \le 0\\
\Leftrightarrow x \le \dfrac{5}{2}\\
Vậy\,x \le \dfrac{5}{2}
\end{array}$
