Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
2{x^2} - 7x + 5 \ge 0\\
1 - x \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {2x - 5} \right)\left( {x - 1} \right) \ge 0\\
x \le 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge \dfrac{5}{2}\\
x \le 1
\end{array} \right.\\
x \le 1
\end{array} \right. \Rightarrow x \le 1\\
\sqrt {2{x^2} - 7x + 5} \le 1 - x\\
\Rightarrow 2{x^2} - 7x + 5 \le {x^2} - 2x + 1\\
\Rightarrow {x^2} - 5x + 4 \le 0\\
\Rightarrow \left( {x - 1} \right)\left( {x - 4} \right) \le 0\\
\Rightarrow 1 \le x \le 4\\
DO:x \le 1\\
\Rightarrow x = 1\\
Vay\,x = 1\\
b)Dkxd:9 - {x^2} \ge 0\\
\Rightarrow {x^2} \le 9\\
\Rightarrow - 3 \le x \le 3\\
3 \ge x - 2\sqrt {9 - {x^2}} \\
\Rightarrow 2\sqrt {9 - {x^2}} \ge x - 3\\
\Rightarrow 4\left( {9 - {x^2}} \right) \ge {x^2} - 6x + 9\\
\Rightarrow 5{x^2} - 6x - 27 \le 0\\
\Rightarrow \left( {5x + 9} \right)\left( {x - 3} \right) \le 0\\
\Rightarrow \dfrac{{ - 9}}{5} \le x \le 3\\
Do: - 3 \le x \le 3\\
Vay\, - \dfrac{9}{5} \le x \le 3
\end{array}$
