Đáp án:
$\begin{array}{l}
1)A = \dfrac{{\sqrt {2009} + 2009}}{{1 + \sqrt {2009} }} - \sqrt {{{\left( {\sqrt {2009} - 9} \right)}^2}} \\
= \dfrac{{\sqrt {2009} \left( {1 + \sqrt {2009} } \right)}}{{1 + \sqrt {2009} }} - \left| {\sqrt {2009} - 9} \right|\\
= \sqrt {2009} - \left( {\sqrt {2009} - 9} \right)\\
= 9\\
2)\sqrt {16{{\left( {x - 2} \right)}^2}} - 8 = 0\\
\Rightarrow 4\left| {x - 2} \right| = 8\\
\Rightarrow \left| {x - 2} \right| = 2\\
\Rightarrow \left[ \begin{array}{l}
x - 2 = 2\\
x - 2 = - 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 4\\
x = 0
\end{array} \right.\\
\text{vậy}\,x = 4;x = 0\\
3)a)Dkxd:x \ge 0;x \ne 1\\
b)A = \dfrac{1}{{\sqrt x + 1}} + \dfrac{{3 - \sqrt x }}{{x - 1}}\\
= \dfrac{1}{{\sqrt x + 1}} + \dfrac{{3 - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x - 1 + 3 - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{2}{{x - 1}}\\
c)B = A.\left( {\sqrt x - 1} \right)\\
= \dfrac{2}{{x - 1}}.\left( {\sqrt x - 1} \right)\\
= \dfrac{2}{{\sqrt x + 1}}\\
Do:\sqrt x \ge 0\\
\Rightarrow \sqrt x + 1 \ge 1\\
\Rightarrow \dfrac{1}{{\sqrt x + 1}} \le 1\\
\Rightarrow \dfrac{2}{{\sqrt x + 1}} \le 2\\
\Rightarrow B \le 2\\
\Rightarrow GTLN:B = 2\,khi:x = 0
\end{array}$
