Đáp án:
$\begin{array}{l}
4)a)Dkxd:x \ge 0;x \ne 1\\
P = 1:\left( {\dfrac{{x + 2}}{{x\sqrt x - 1}} + \dfrac{{\sqrt x + 1}}{{x + \sqrt x + 1}} - \dfrac{{\sqrt x + 1}}{{x - 1}}} \right)\\
= 1:\left( {\dfrac{{x + 2}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} + \dfrac{{\sqrt x + 1}}{{x + \sqrt x + 1}} - \dfrac{1}{{\sqrt x - 1}}} \right)\\
= 1:\dfrac{{x + 2 + \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{x + 2 + x - 1 - x - \sqrt x - 1}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{x - \sqrt x }}\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x }}\\
b)P - 3\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x }} - 3\\
= \dfrac{{x + \sqrt x + 1 - 3\sqrt x }}{{\sqrt x }}\\
= \dfrac{{x - 2\sqrt x + 1}}{{\sqrt x }} = \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }} > 0\\
\Rightarrow P > 3\\
5)a)Dkxd:x \ge 0;x \ne 1\\
A = \dfrac{{x\sqrt x + 26\sqrt x - 19}}{{x + 2\sqrt x - 3}} - \dfrac{{2\sqrt x }}{{\sqrt x - 1}} + \dfrac{{\sqrt x - 3}}{{\sqrt x + 3}}\\
= \dfrac{{x\sqrt x + 26\sqrt x - 19 - 2\sqrt x \left( {\sqrt x + 3} \right) + \left( {\sqrt x - 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x\sqrt x + 26\sqrt x - 19 - 2x - 6\sqrt x + x - 4\sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x\sqrt x - x + 16\sqrt x - 16}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {x + 16} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x + 16}}{{\sqrt x + 3}}\\
b)P = \dfrac{{x + 16}}{{\sqrt x + 3}} = \dfrac{{x + 3\sqrt x - 3\sqrt x - 9 + 25}}{{\sqrt x + 3}}\\
= \sqrt x - 3 + \dfrac{{25}}{{\sqrt x + 3}}\\
= \sqrt x + 3 + \dfrac{{25}}{{\sqrt x + 3}} - 6\\
Theo\,Co - si\\
\left( {\sqrt x + 3} \right) + \dfrac{{25}}{{\sqrt x + 3}} \ge 2\sqrt {\left( {\sqrt x + 3} \right).\dfrac{{25}}{{\sqrt x + 3}}} = 10\\
\Rightarrow P \ge 10 - 6 = 4\\
\Rightarrow P \ge 4\\
\Rightarrow GTNN:P = 4\,\\
Khi:\left( {\sqrt x + 3} \right) = \dfrac{{25}}{{\sqrt x + 3}}\\
\Rightarrow \sqrt x + 3 = 5\\
\Rightarrow \sqrt x = 2\\
\Rightarrow x = 4\left( {tmdk} \right)
\end{array}$
