Đáp án:
$\begin{array}{l}
a)x = \sqrt {3 + 2\sqrt 2 } + \sqrt {11 - 6\sqrt 2 } \left( {tmdk} \right)\\
= \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} + \sqrt {{{\left( {3 - \sqrt 2 } \right)}^2}} \\
= \sqrt 2 + 1 + 3 - \sqrt 2 \\
= 4\\
\Leftrightarrow \sqrt x = 2\\
B = \dfrac{{\sqrt x + 5}}{{\sqrt x - 3}} = \dfrac{{2 + 5}}{{2 - 3}} = - 7\\
b)P = \dfrac{A}{B}\\
= \left( {\dfrac{4}{{\sqrt x + 3}} + \dfrac{{2x - \sqrt x - 13}}{{x - 9}} - \dfrac{{\sqrt x }}{{\sqrt x - 3}}} \right)\\
:\dfrac{{\sqrt x + 5}}{{\sqrt x - 3}}\\
= \dfrac{{4\left( {\sqrt x - 3} \right) + 2x - \sqrt x - 13 - \sqrt x \left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
.\dfrac{{\sqrt x - 3}}{{\sqrt x + 5}}\\
= \dfrac{{4\sqrt x - 12 + 2x - \sqrt x - 13 - x - 3\sqrt x }}{{\sqrt x + 3}}.\dfrac{1}{{\sqrt x + 5}}\\
= \dfrac{{x - 25}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x + 5} \right)}}\\
= \dfrac{{\sqrt x - 5}}{{\sqrt x + 3}}\\
c) - {P^2} + 3P - 5\\
= - \left( {{P^2} - 3P} \right) - 5\\
= - \left( {{P^2} - 2.P.\dfrac{3}{2} + \dfrac{9}{4}} \right) + \dfrac{9}{4} - 5\\
= - {\left( {P - \dfrac{3}{2}} \right)^2} - \dfrac{{11}}{4} \le \dfrac{{ - 11}}{4}\\
\Leftrightarrow GTLN:\left( { - {P^2} + 3P - 5\,} \right) = - \dfrac{{11}}{4}\\
Khi:P = \dfrac{3}{2}\\
\Leftrightarrow \dfrac{{\sqrt x - 5}}{{\sqrt x + 3}} = \dfrac{3}{2}\\
\Leftrightarrow 3\sqrt x + 9 = 2\sqrt x - 10\\
\Leftrightarrow \sqrt x = - 19\left( {ktm} \right)
\end{array}$
Vậy ko có giá trị của x để biểu thức đạt GTLN
