Đáp án:
$\begin{array}{l}
1)a)A = \sqrt {121} - \sqrt {25} .\sqrt 4 \\
= 11 - 5.2\\
= 1\\
b)B = \sqrt {18} - 2\sqrt {50} + 3\sqrt 8 \\
= 3\sqrt 2 - 2.5\sqrt 2 + 3.2\sqrt 2 \\
= 3\sqrt 2 - 10\sqrt 2 + 6\sqrt 2 \\
= - \sqrt 2 \\
c)C = \left( {5 - \dfrac{{7 - \sqrt 7 }}{{\sqrt 7 - 1}}} \right).\left( {5 + \dfrac{{\sqrt 7 + 7}}{{\sqrt 7 + 1}}} \right)\\
= \left( {5 - \sqrt 7 } \right).\left( {5 + \sqrt 7 } \right)\\
= {5^2} - 7\\
= 25 - 7\\
= 18\\
2)a)\sqrt {x - 2011} = 3\left( {dk:x \ge 2011} \right)\\
\Rightarrow x - 2011 = 9\\
\Rightarrow x = 2020\left( {tmdk} \right)\\
Vậy\,x = 2020\\
b)Dkxd:x \ge - 3\\
3\sqrt {4x + 12} - \sqrt {9x + 27} + \sqrt {x + 3} = 8\\
\Rightarrow 3.2\sqrt {x + 3} - 3.\sqrt {x + 3} + \sqrt {x + 3} = 8\\
\Rightarrow 4\sqrt {x + 3} = 8\\
\Rightarrow \sqrt {x + 3} = 2\\
\Rightarrow x + 3 = 4\\
\Rightarrow x = 1\left( {tm} \right)\\
Vậy\,x = 1\\
d)\sqrt {4{x^2} - 12x + 9} = 5\\
\Rightarrow \sqrt {{{\left( {2x - 3} \right)}^2}} = 5\\
\Rightarrow \left| {2x - 3} \right| = 5\\
\Rightarrow \left[ \begin{array}{l}
2x - 3 = 5\\
2x - 3 = - 5
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 4\\
x = - 1
\end{array} \right.\\
Vậy\,x = 4;x = - 1\\
3)a)x \ge 0;x \ne 9\\
A = \dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 9}}{{x - 9}}\\
= \dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - 3x - 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - 3\sqrt x - 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - 3\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - 3}}{{\sqrt x - 3}}\\
= \dfrac{3}{{3 - \sqrt x }}\\
b)A = - 2 \Rightarrow \dfrac{3}{{3 - \sqrt x }} = - 2\\
\Rightarrow 3 - \sqrt x = - \dfrac{3}{2} \Rightarrow \sqrt x = \dfrac{9}{2}\\
\Rightarrow x = \dfrac{{81}}{4}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{{81}}{4}\\
c)P > 0\\
\Rightarrow \dfrac{3}{{3 - \sqrt x }} > 0 \Rightarrow 3 - \sqrt x > 0\\
\Rightarrow \sqrt x < 3 \Rightarrow x < 9\\
Vậy\,0 \le x < 9
\end{array}$
