Đáp án:
\(Min = - \dfrac{1}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
P = \left[ {\dfrac{{x + \sqrt x - x - 2}}{{\sqrt x + 1}}} \right]:\left[ {\dfrac{{x - \sqrt x + \sqrt x - 4}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}} \right]\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}.\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{x - 4}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}} = \dfrac{{\sqrt x + 2 - 3}}{{\sqrt x + 2}} = 1 - \dfrac{3}{{\sqrt x + 2}}\\
Do:x \ge 0\\
\to \sqrt x \ge 0\\
\to \sqrt x + 2 \ge 2\\
\to \dfrac{3}{{\sqrt x + 2}} \le \dfrac{3}{2}\\
\to - \dfrac{3}{{\sqrt x + 2}} \ge - \dfrac{3}{2}\\
\to 1 - \dfrac{3}{{\sqrt x + 2}} \ge - \dfrac{1}{2}\\
\to Min = - \dfrac{1}{2}\\
\Leftrightarrow x = 0
\end{array}\)
