Đáp án:
$\begin{array}{l}
B2)a)\\
P = \left( {\dfrac{{x + 2}}{{x\sqrt x - 1}} + \dfrac{{\sqrt x }}{{x + \sqrt x + 1}} + \dfrac{1}{{1 - \sqrt x }}} \right):\dfrac{{\sqrt x - 1}}{2}\\
= \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)}^2}.\left( {x + \sqrt x + 1} \right)}}.2\\
= \dfrac{{x - 2\sqrt x + 1}}{{{{\left( {\sqrt x - 1} \right)}^2}.\left( {x + \sqrt x + 1} \right)}}.2\\
= \dfrac{2}{{x + \sqrt x + 1}}\\
b)Dkx{\rm{d}}:x > 0;x \ne 1\\
P = \dfrac{2}{7}\\
\Rightarrow \dfrac{2}{{x + \sqrt x + 1}} = \dfrac{2}{7}\\
\Rightarrow x + \sqrt x + 1 = 7\\
\Rightarrow x + \sqrt x - 6 = 0\\
\Rightarrow \left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right) = 0\\
\Rightarrow \sqrt x = 2\\
\Rightarrow x = 4\left( {tmdk} \right)\\
c)P = \dfrac{2}{{x + \sqrt x + 1}}\\
\Rightarrow {P^2} = \dfrac{4}{{{{\left( {x + \sqrt x + 1} \right)}^2}}}\\
\Rightarrow {P^2} - P = \dfrac{4}{{{{\left( {x + \sqrt x + 1} \right)}^2}}} - \dfrac{2}{{x + \sqrt x + 1}}\\
= \dfrac{{4 - 2\left( {x + \sqrt x + 1} \right)}}{{{{\left( {x + \sqrt x + 1} \right)}^2}}}\\
= \dfrac{{ - 2x - 2\sqrt x + 2}}{{{{\left( {x + \sqrt x + 1} \right)}^2}}}\\
= \dfrac{{ - 2\left( {x + \sqrt x - 1} \right)}}{{{{\left( {x + \sqrt x + 1} \right)}^2}}} > 0\\
\Rightarrow {P^2} > P\\
B3)\\
a)P = \left( {\dfrac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}} - \dfrac{{\sqrt x - 2}}{{x - 1}}} \right).\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \left( {\dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right) - \left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}} \right).\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{x + \sqrt x - 2 - \left( {x - \sqrt x - 2} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right).\sqrt x }}\\
= \dfrac{{2\sqrt x }}{{\sqrt x \left( {x - 1} \right)}}\\
= \dfrac{2}{{x - 1}}\\
b)P = \dfrac{2}{{x - 1}} \in Z\\
\Rightarrow \left( {x - 1} \right) \in \left\{ { - 2; - 1;1;2} \right\}\\
\Rightarrow x \in \left\{ { - 1;0;2;3} \right\}\\
Do:x > 0;x \ne 1\\
\Rightarrow x = 3
\end{array}$
