Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
9 - x \ge 0\\
x - 1 \ge 0
\end{array} \right. \Rightarrow 1 \le x \le 9\\
A = \sqrt {9 - x} + \sqrt {x - 1} \\
Theo\,Bunhia:\\
{A^2} = {\left( {\sqrt {9 - x} + \sqrt {x - 1} } \right)^2} \le \left( {1 + 1} \right).\left( {9 - x + x - 1} \right)\\
\Rightarrow {A^2} \le 2.10 = 20\\
\Rightarrow - \sqrt {20} \le A \le \sqrt {20} \\
\Rightarrow A \le 2\sqrt 5 \\
\Rightarrow GTLN:A = 2\sqrt 5 \,\\
Khi:\sqrt {9 - x} = \sqrt {x - 1} \\
\Rightarrow 9 - x = x - 1\\
\Rightarrow x = 5\left( {tmdk} \right)\\
b)Dkxd:x \ge 1\\
\sqrt {x + 2\sqrt {x - 1} } + \sqrt {x - 2\sqrt {x - 1} } = 2\\
\Rightarrow \sqrt {x - 1 + 2\sqrt {x - 1} + 1} + \sqrt {x - 1 - 2\sqrt {x - 1} + 1} = 2\\
\Rightarrow \sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} = 2\\
\Rightarrow \sqrt {x - 1} + 1 + \left| {\sqrt {x - 1} - 1} \right| = 2\\
+ )Khi:\sqrt {x - 1} - 1 \ge 0\\
\Rightarrow \sqrt {x - 1} \ge 1\\
\Rightarrow x \ge 2\\
\Rightarrow \sqrt {x - 1} + 1 + \sqrt {x - 1} - 1 = 2\\
\Rightarrow \sqrt {x - 1} = 1\\
\Rightarrow x = 2\left( {tmdk} \right)\\
+ Khi:\sqrt {x - 1} < 1\\
\Rightarrow 1 \le x < 2\\
\Rightarrow \sqrt {x - 1} + 1 + 1 - \sqrt {x - 1} = 2\\
\Rightarrow 2 = 2\left( {tm} \right)\\
\text{Vậy}\,1 \le x \le 2
\end{array}$
