\(\begin{array}{l}
A = - \dfrac{x + 7}{\sqrt x + 1}\\
B = \dfrac{\sqrt x - 2}{\sqrt x + 3} - \dfrac{\sqrt x}{3 - \sqrt x} - \dfrac{x + 9}{x - 9}\qquad (x \geqslant 0;\ x \ne 9)\\
a)\quad \text{Khi $x = \dfrac14$ ta được:}\\
\quad A = - \dfrac{\dfrac14 + 7}{\sqrt{\dfrac14} + 1}\\
\to A = - \dfrac{\dfrac{29}{4}}{\dfrac12 + 1}\\
\to A = - \dfrac{29}{4}\cdot \dfrac23\\
\to A = - \dfrac{29}{6}\\
b)\quad B = \dfrac{\sqrt x - 2}{\sqrt x + 3}+\dfrac{\sqrt x}{\sqrt x - 3} - \dfrac{x + 9}{\left(\sqrt x + 3\right)\left(\sqrt x - 3\right)}\\
\to B= \dfrac{\left(\sqrt x - 2\right)\left(\sqrt x - 3\right)}{\left(\sqrt x + 3\right)\left(\sqrt x - 3\right)} + \dfrac{\sqrt x\left(\sqrt x + 3\right)}{\left(\sqrt x + 3\right)\left(\sqrt x - 3\right)} - \dfrac{x + 9}{\left(\sqrt x + 3\right)\left(\sqrt x - 3\right)}\\
\to B= \dfrac{\left(x - 5\sqrt x + 6\right) + \left(x + 3\sqrt x\right) - (x + 9)}{\left(\sqrt x + 3\right)\left(\sqrt x - 3\right)}\\
\to B= \dfrac{x - 2\sqrt x - 3}{\left(\sqrt x + 3\right)\left(\sqrt x - 3\right)}\\
\to B = \dfrac{\left(\sqrt x + 1\right)\left(\sqrt x - 3\right)}{\left(\sqrt x + 3\right)\left(\sqrt x - 3\right)}\\
\to B = \dfrac{\sqrt x + 1}{\sqrt x + 3}\\
c)\quad P = AB\\
\to P = \left(- \dfrac{x + 7}{\sqrt x + 1}\right)\cdot \dfrac{\sqrt x + 1}{\sqrt x + 3}\\
\to P = - \dfrac{x + 7}{\sqrt x + 3}\\
\to P = - \sqrt x + 3 - \dfrac{2}{\sqrt x + 3}\\
\to P = - \left(\sqrt x + 3 + \dfrac{2}{\sqrt x + 3}\right)+ 6\\
\text{Áp dụng bất đẳng thức AM - GM ta được:}\\
\quad \sqrt x + 3 + \dfrac{2}{\sqrt x + 3} \geqslant 2\sqrt{\left(\sqrt x + 3\right)\cdot \dfrac{2}{\sqrt x + 3}}\\
\Leftrightarrow \sqrt x + 3 + \dfrac{2}{\sqrt x + 3} \geqslant 2\sqrt2\\
\Leftrightarrow - \left(\sqrt x + 3 + \dfrac{2}{\sqrt x + 3}\right) \leqslant -2\sqrt2\\
\Leftrightarrow - \left(\sqrt x + 3 + \dfrac{2}{\sqrt x + 3}\right) + 6 \leqslant 6 - 2\sqrt2\\
\Leftrightarrow P \leqslant 6 - 2\sqrt2\\
\text{Dấu = xảy ra}\ \sqrt x + 3 = \dfrac{2}{\sqrt x + 3} \Leftrightarrow x = 11 - 6\sqrt2
\end{array}\)
