Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
P = \dfrac{{2x + 2}}{{\sqrt x }} + \dfrac{{x\sqrt x - 1}}{{x - \sqrt x }} - \dfrac{{x\sqrt x + 1}}{{x + \sqrt x }}\\
= \dfrac{{2x + 2}}{{\sqrt x }} + \dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}} - \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{2x + 2}}{{\sqrt x }} + \dfrac{{x + \sqrt x + 1}}{{\sqrt x }} - \dfrac{{x - \sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{\left( {2x + 2} \right) + \left( {x + \sqrt x + 1} \right) - \left( {x - \sqrt x + 1} \right)}}{{\sqrt x }}\\
= \dfrac{{2x + 2\sqrt x + 2}}{{\sqrt x }}\\
b,\\
P = \dfrac{{2x + 2\sqrt x + 2}}{{\sqrt x }} = 2\sqrt x + 2 + \dfrac{2}{{\sqrt x }}\\
= 2.\left( {\sqrt x + \dfrac{1}{{\sqrt x }}} \right) + 2 \ge 2.2\sqrt {\sqrt x .\dfrac{1}{{\sqrt x }}} + 2 = 4 + 2 = 6 > 5\\
\Rightarrow P > 5\\
c,\\
P = \dfrac{{2x + 2\sqrt x + 2}}{{\sqrt x }} > 0,\,\,\,\,\forall x > 0 \Rightarrow \dfrac{8}{P} > 0\\
P \ge 6 \Rightarrow \dfrac{8}{P} \le \dfrac{8}{6}\\
\Rightarrow 0 < \dfrac{8}{P} \le \dfrac{8}{6}\\
\dfrac{8}{P} \in Z \Rightarrow \dfrac{8}{P} = 1
\end{array}\)
