Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
DKXD:\,\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
a,\\
P = \left( {\sqrt x - \dfrac{{x + 2}}{{\sqrt x + 1}}} \right):\left( {\dfrac{{\sqrt x }}{{\sqrt x + 1}} - \dfrac{{\sqrt x - 4}}{{1 - x}}} \right)\\
= \dfrac{{\sqrt x .\left( {\sqrt x + 1} \right) - \left( {x + 2} \right)}}{{\sqrt x + 1}}:\left( {\dfrac{{\sqrt x }}{{\sqrt x + 1}} + \dfrac{{\sqrt x - 4}}{{x - 1}}} \right)\\
= \dfrac{{\left( {x + \sqrt x } \right) - \left( {x + 2} \right)}}{{\sqrt x + 1}}:\left( {\dfrac{{\sqrt x }}{{\sqrt x + 1}} + \dfrac{{\sqrt x - 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right)\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}:\dfrac{{\sqrt x .\left( {\sqrt x - 1} \right) + \left( {\sqrt x - 4} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}:\dfrac{{x - 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}}\\
b,\\
P < 0 \Leftrightarrow \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}} < 0 \Leftrightarrow \sqrt x - 1 < 0 \Leftrightarrow \sqrt x < 1 \Leftrightarrow 0 \le x < 1\\
c,\\
P = \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}} = \dfrac{{\left( {\sqrt x + 2} \right) - 3}}{{\sqrt x + 2}} = 1 - \dfrac{3}{{\sqrt x + 2}}\\
\sqrt x + 2 \ge 2,\,\,\,\forall x \Rightarrow \dfrac{3}{{\sqrt x + 2}} \le \dfrac{3}{2},\,\,\,\forall x\\
\Rightarrow P = 1 - \dfrac{3}{{\sqrt x + 2}} \ge 1 - \dfrac{3}{2} = - \dfrac{1}{2},\,\,\,\forall x\\
\Rightarrow {P_{\min }} = - \dfrac{1}{2} \Leftrightarrow x = 0
\end{array}\)
