Đáp án:
$\begin{array}{l}
b)x > 0;x \ne 1\\
P = \dfrac{{1 + x}}{{\sqrt x }}\\
P - 2 = \dfrac{{1 + x}}{{\sqrt x }} - 2\\
= \dfrac{{1 + x - 2\sqrt x }}{{\sqrt x }}\\
= \dfrac{{x - 2\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }}\\
Do:x \ne 1;x > 0 \Rightarrow \left\{ \begin{array}{l}
{\left( {\sqrt x - 1} \right)^2} > 0\\
\sqrt x > 0
\end{array} \right.\\
\Rightarrow \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }} > 0\\
\Rightarrow P - 2 > 0\\
\Rightarrow P > 2
\end{array}$
