Đáp án:
$\begin{array}{l}
a)DKxd:x \ge 0;x \ne 1\\
P = \left( {\frac{{\sqrt x - 2}}{{x - 1}} - \frac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}}} \right).{\left( {\frac{{1 - x}}{{\sqrt 2 }}} \right)^2}\\
= \frac{{\left( {\sqrt x - 2} \right).\left( {x + 2\sqrt x + 1} \right) - \left( {\sqrt x + 2} \right).\left( {x - 1} \right)}}{{\left( {x - 1} \right).{{\left( {\sqrt x + 1} \right)}^2}}}.\frac{{{{\left( {x - 1} \right)}^2}}}{2}\\
= \frac{{x\sqrt x + 2 + \sqrt x - 2x - 4\sqrt x - 2 - \left( {x\sqrt x - \sqrt x + 2x - 2} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}}}.\frac{{x - 1}}{2}\\
= \frac{{ - 4x - 2\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}}.\frac{{x - 1}}{2}\\
= \frac{{2x + \sqrt x - 1}}{{{{\left( {\sqrt x + 1} \right)}^2}}}.\left( {x - 1} \right)\\
= \frac{{\left( {\sqrt x + 1} \right).\left( {2\sqrt x - 1} \right)\left( {x - 1} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}}}\\
= \frac{{\left( {2\sqrt x - 1} \right).\left( {x - 1} \right)}}{{\sqrt x + 1}}\\
b)Dkxd:x \ge 0;x \ne 1\\
P = - 2\\
\Rightarrow \frac{{\left( {2\sqrt x - 1} \right).\left( {x - 1} \right)}}{{\sqrt x + 1}} = - 2\\
\Rightarrow 2x\sqrt x - 2\sqrt x - x + 1 = - 2\sqrt x - 2\\
\Rightarrow 2x\sqrt x - x + 3 = 0\\
\Rightarrow 2x\sqrt x + 2x - 3x - 3\sqrt x + 3\sqrt x + 3 = 0\\
\Rightarrow \left( {\sqrt x + 1} \right)\left( {2x - 3\sqrt x + 3} \right) = 0\\
\Rightarrow \sqrt x = - 1\left( {ktm} \right)\\
Vậy\,ko\,có\,x\,để\,P = - 2
\end{array}$
