Đáp án:
$\begin{array}{l}
Dkxd:x > 0\\
P = \left( {\dfrac{1}{{\sqrt x }} + \dfrac{{\sqrt x }}{{\sqrt x + 1}}} \right):\dfrac{{\sqrt x }}{{x + \sqrt x }}\\
= \dfrac{{\sqrt x + 1 + \sqrt x .\sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x }}\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}.\left( {\sqrt x + 1} \right)\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x }}\\
P = - 1\\
\Rightarrow \dfrac{{x + \sqrt x + 1}}{{\sqrt x }} = - 1\\
\Rightarrow x + \sqrt x + 1 = - \sqrt x \\
\Rightarrow x + 2\sqrt x + 1 = 0\\
\Rightarrow {\left( {\sqrt x + 1} \right)^2} = 0\\
\Rightarrow \sqrt x = - 1\left( {vn} \right)
\end{array}$
Vậy ko có x thỏa mãn P=-1
