Giải thích các bước giải:
a) ĐKXĐ: $x \ge 0;x \ne 1$
Ta có:
$\begin{array}{l}
P = \dfrac{{10\sqrt x }}{{x + 3\sqrt x - 4}} - \dfrac{{2\sqrt x - 3}}{{4 + \sqrt x }} + \dfrac{{\sqrt x + 1}}{{1 - \sqrt x }}\\
= \dfrac{{10\sqrt x }}{{\left( {\sqrt x + 4} \right)\left( {\sqrt x - 1} \right)}} - \dfrac{{2\sqrt x - 3}}{{\sqrt x + 4}} - \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{10\sqrt x - \left( {2\sqrt x - 3} \right)\left( {\sqrt x - 1} \right) - \left( {\sqrt x + 1} \right)\left( {\sqrt x + 4} \right)}}{{\left( {\sqrt x + 4} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{ - 3x + 10\sqrt x - 7}}{{\left( {\sqrt x + 4} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( { - 3\sqrt x + 7} \right)}}{{\left( {\sqrt x + 4} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{ - 3\sqrt x + 7}}{{\sqrt x + 4}}
\end{array}$
Vậy $P = \dfrac{{ - 3\sqrt x + 7}}{{\sqrt x + 4}}$ với $x \ge 0;x \ne 1$
b) Ta có:
$\begin{array}{l}
P = \dfrac{{ - 3\sqrt x + 7}}{{\sqrt x + 4}}\\
= \dfrac{{ - 3\left( {\sqrt x + 4} \right) + 5}}{{\sqrt x + 4}}\\
= - 3 + \dfrac{5}{{\sqrt x + 4}}\\
> - 3\left( {do:\dfrac{5}{{\sqrt x + 4}} > 0,\forall x \ge 0;x \ne 1} \right)
\end{array}$
c) Ta có:
$\begin{array}{l}
P - \dfrac{7}{4} = \dfrac{{ - 3\sqrt x + 7}}{{\sqrt x + 4}} - \dfrac{7}{4}\\
= \dfrac{{4\left( { - 3\sqrt x + 7} \right) - 7\left( {\sqrt x + 4} \right)}}{{4\left( {\sqrt x + 4} \right)}}\\
= \dfrac{{ - 19\sqrt x }}{{4\left( {\sqrt x + 4} \right)}}\\
\le 0,\forall x \ge 0,x \ne 1\\
\Rightarrow P \le \dfrac{7}{4}
\end{array}$
Dấu bằng xảy ra
$ \Leftrightarrow \sqrt x = 0 \Leftrightarrow x = 0$
Vậy $MaxP = \dfrac{7}{4} \Leftrightarrow x = 0$
