Đáp án:
$\begin{array}{l}
a)Dkxd:x > 0;x \ne 1\\
P = \dfrac{{x\sqrt x }}{{x - \sqrt x }} - \dfrac{{x\sqrt x + 1}}{{x + \sqrt x }} + \dfrac{{x + 1}}{{\sqrt x }}\\
= \dfrac{{x\sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}} - \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x + 1} \right)}} + \dfrac{{x + 1}}{{\sqrt x }}\\
= \dfrac{x}{{\sqrt x - 1}} - \dfrac{{x - \sqrt x + 1}}{{\sqrt x }} + \dfrac{{x + 1}}{{\sqrt x }}\\
= \dfrac{x}{{\sqrt x - 1}} + \dfrac{{ - x + \sqrt x - 1 + x + 1}}{{\sqrt x }}\\
= \dfrac{x}{{\sqrt x - 1}} + \dfrac{{\sqrt x }}{{\sqrt x }}\\
= \dfrac{x}{{\sqrt x - 1}} + 1\\
= \dfrac{{x + \sqrt x - 1}}{{\sqrt x - 1}}\\
b)P = \dfrac{9}{2}\\
\Rightarrow \dfrac{{x + \sqrt x - 1}}{{\sqrt x - 1}} = \dfrac{9}{2}\\
\Rightarrow 2x + 2\sqrt x - 2 = 9\sqrt x - 9\\
\Rightarrow 2x - 7\sqrt x + 7 = 0\\
\Rightarrow x - \dfrac{7}{2}\sqrt x + \dfrac{7}{2} = 0\\
\Rightarrow x - 2.\dfrac{7}{4}\sqrt x + \dfrac{{49}}{{16}} + \dfrac{7}{{16}} = 0\\
\Rightarrow {\left( {\sqrt x - \dfrac{7}{4}} \right)^2} + \dfrac{7}{{16}} = 0\left( {vn} \right)
\end{array}$
Vậy ko có giá trị của x để P=9/2
