Đáp án:
\(P \le 1\forall x > 0\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x > 0\\
P = \left[ {\dfrac{{x + 8\sqrt x + 8 - \left( {x + 4\sqrt x + 4} \right)}}{{\sqrt x \left( {\sqrt x + 2} \right)}}} \right]:\left[ {\dfrac{{x + \sqrt x + 3 + \sqrt x + 2}}{{\sqrt x \left( {\sqrt x + 2} \right)}}} \right]\\
= \dfrac{{4\sqrt x + 4}}{{\sqrt x \left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{x + 2\sqrt x + 5}}\\
= \dfrac{{4\sqrt x + 4}}{{x + 2\sqrt x + 5}}\\
P \le 1\\
\to \dfrac{{4\sqrt x + 4}}{{x + 2\sqrt x + 5}} \le 1\\
\to \dfrac{{4\sqrt x + 4 - x - 2\sqrt x - 5}}{{x + 2\sqrt x + 5}} \le 0\\
\to \dfrac{{ - x + 2\sqrt x - 1}}{{x + 2\sqrt x + 5}} \le 0\\
\to - x + 2\sqrt x - 1 \le 0\left( {do:x + 2\sqrt x + 5 > 0\forall x > 0} \right)\\
\to - {\left( {\sqrt x - 1} \right)^2} \le 0\\
\to {\left( {\sqrt x - 1} \right)^2} \ge 0\left( {ld} \right)\forall x > 0\\
\to P \le 1\forall x > 0
\end{array}\)
