Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 1\\
b.P = \left[ {\frac{{x - \sqrt x - 2 - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right].\frac{{{{\left( {1 - x} \right)}^2}}}{2}\\
= \frac{{x - \sqrt x - 2 - x - \sqrt x + 2}}{{\left( {x - 1} \right)\left( {\sqrt x + 1} \right)}}.\frac{{{{\left( {x - 1} \right)}^2}}}{2}\\
= \frac{{ - 2\sqrt x }}{{\left( {x - 1} \right)\left( {\sqrt x + 1} \right)}}.\frac{{{{\left( {x - 1} \right)}^2}}}{2}\\
= \frac{{ - \sqrt x }}{{\sqrt x - 1}}\\
c.P > 0\\
\to \frac{{ - \sqrt x }}{{\sqrt x - 1}} > 0\\
\to \frac{{\sqrt x }}{{\sqrt x - 1}} < 0\\
\to \sqrt x - 1 < 0\left( {do\sqrt x \ge 0\forall x \ge 0} \right)\\
\to \sqrt x < 1\\
\to 0 \le x < 1\\
d.P = \frac{{ - \sqrt x }}{{\sqrt x - 1}} = \frac{{ - \left( {\sqrt x - 1} \right) - 1}}{{\sqrt x - 1}} = - 1 - \frac{1}{{\sqrt x - 1}}\\
= - \left( {1 + \frac{1}{{\sqrt x - 1}}} \right)\\
Do:\frac{1}{{\sqrt x - 1}} > 0\forall x \ge 0;x \ne 1\\
\to 1 + \frac{1}{{\sqrt x - 1}} > 1\\
\to - \left( {1 + \frac{1}{{\sqrt x - 1}}} \right) < - 1\\
\to Max = - 2\\
\Leftrightarrow \frac{1}{{\sqrt x - 1}} = 1\\
\to \sqrt x - 1 = 1\\
\to x = 4
\end{array}\)
