Đáp án:

$\begin{array}{l}
a)Dkxd:a > 0;a \ne 9\\
P = \left( {\frac{{\sqrt a }}{{3 + \sqrt a }} + \frac{{a + 9}}{{9 - a}}} \right):\left( {\frac{{3\sqrt a  + 1}}{{a - 3\sqrt a }} - \frac{1}{{\sqrt a }}} \right)\\
 = \left( {\frac{{\sqrt a }}{{3 + \sqrt a }} - \frac{{a + 9}}{{\left( {\sqrt a  + 3} \right)\left( {\sqrt a  - 3} \right)}}} \right):\left( {\frac{{3\sqrt a  + 1}}{{\sqrt a \left( {\sqrt a  - 3} \right)}} - \frac{1}{{\sqrt a }}} \right)\\
 = \frac{{\sqrt a \left( {\sqrt a  - 3} \right) - a - 9}}{{\left( {\sqrt a  + 3} \right)\left( {\sqrt a  - 3} \right)}}:\frac{{3\sqrt a  + 1 - \sqrt a  + 3}}{{\sqrt a \left( {\sqrt a  - 3} \right)}}\\
 = \frac{{a - 3\sqrt a  - a - 9}}{{\left( {\sqrt a  + 3} \right)\left( {\sqrt a  - 3} \right)}}:\frac{{2\sqrt a  + 4}}{{\sqrt a \left( {\sqrt a  - 3} \right)}}\\
 = \frac{{ - 3\left( {\sqrt a  + 3} \right)}}{{\left( {\sqrt a  + 3} \right)\left( {\sqrt a  - 3} \right)}}.\frac{{\sqrt a \left( {\sqrt a  - 3} \right)}}{{2\sqrt a  + 4}}\\
 = \frac{{ - 3\sqrt a }}{{2\sqrt a  + 4}}\\
b)a = 3 - 2\sqrt 2 \left( {tm} \right)\\
 \Rightarrow a = 2 - 2\sqrt 2  + 1 = {\left( {\sqrt 2  - 1} \right)^2}\\
 \Rightarrow \sqrt a  = \sqrt 2  - 1\\
 \Rightarrow P = \frac{{ - 3\sqrt a }}{{2\sqrt a  + 4}} = \frac{{ - 3\left( {\sqrt 2  - 1} \right)}}{{2\left( {\sqrt 2  - 1} \right) + 4}}\\
 = \frac{{ - 3\sqrt 2  + 3}}{{2\sqrt 2  + 2}}\\
 = \frac{{ - 3\left( {\sqrt 2  - 1} \right).\left( {\sqrt 2  - 1} \right)}}{{2\left( {\sqrt 2  + 1} \right)\left( {\sqrt 2  - 1} \right)}}\\
 = \frac{{ - 3{{\left( {\sqrt 2  - 1} \right)}^2}}}{2}\\
 = \frac{{ - 3\left( {3 - 2\sqrt 2 } \right)}}{2}\\
 = \frac{{6\sqrt 2  - 9}}{2}
\end{array}$

Đáp án:

Giải thích các bước giải:

$\begin{gathered}   a)\,P = (\frac{{\sqrt a }}{{3 + \sqrt a }} + \frac{{a + 9}}{{9 - a}}):(\frac{{3\sqrt a  + 1}}{{a - 3\sqrt a }} - \frac{1}{{\sqrt a }}) \hfill \\    = \frac{{\sqrt a (3 - \sqrt a ) + a + 9}}{{(3 + \sqrt a )(3 - \sqrt a )}}:\frac{{3\sqrt a  + 1 - (\sqrt a  - 3)}}{{\sqrt a (\sqrt a  - 3)}} \hfill \\    = \frac{{ - 2a + 3\sqrt a  + a + 9}}{{(3 + \sqrt a )(3 - \sqrt a )}}.\frac{{\sqrt a (\sqrt a  - 3)}}{{2\sqrt a  + 4}} \hfill \\    = \frac{{(a - 3\sqrt a  - 9)\sqrt a }}{{(2\sqrt a  + 4)(\sqrt a  + 3)}} \hfill \\   b)\,Khi\,a = 3 - 2\sqrt 2  \hfill \\    =  > \,a = {\left( {\sqrt 2 } \right)^2} - 2\sqrt 2  + 1 = {\left( {\sqrt 2  - 1} \right)^2} \hfill \\    =  > \,\sqrt a  = \sqrt 2  - 1(do\,\sqrt 2  - 1 > 0) \hfill \\    \hfill \\  \end{gathered} $