\(\begin{array}{l}
9)\,a)\,P = \dfrac{{x + 2 + \sqrt x \left( {\sqrt x - 1} \right) - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{2}{{\sqrt x - 1}}\\
= \dfrac{{x + 2 + x - \sqrt x - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{2}{{\sqrt x - 1}}\\
= \dfrac{{x - 2\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{2}{{\sqrt x - 1}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}.2}}{{{{\left( {\sqrt x - 1} \right)}^2}\left( {x + \sqrt x + 1} \right)}} = \dfrac{2}{{x + \sqrt x + 1}}\\
b)\,x + \sqrt x + 1 > 0\,voi\,moi\,x \ge 0,x \ne 1\\
\Rightarrow P > 0\\
10)\,DK:\,a > 0,a \ne 1\\
Q = \dfrac{{\sqrt a - 1 - \sqrt a - 1 + 2{a^2} + 2}}{{2\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}.\dfrac{{a + 1}}{a}\\
= \dfrac{{2{a^2}}}{{2\left( {{a^2} - 1} \right)}}.\dfrac{{a + 1}}{a}\\
= \dfrac{{a\left( {a + 1} \right)}}{{{a^2} - 1}}
\end{array}\)
Câu 10 em xem lại đề bài nhé
