Đáp án:
\(Min = - \dfrac{1}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
c)B = \dfrac{{2\sqrt x \left( {\sqrt x + 5} \right) - x + 25\sqrt x - \sqrt x \left( {\sqrt x - 5} \right)}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}\\
= \dfrac{{2x + 10\sqrt x - x + 25\sqrt x - x + 5\sqrt x }}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}\\
= \dfrac{{40\sqrt x }}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}\\
P = \dfrac{B}{A} = \dfrac{{2\sqrt x }}{{\sqrt x + 5}}:\dfrac{{40\sqrt x }}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}\\
= \dfrac{{2\sqrt x }}{{\sqrt x + 5}}.\dfrac{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}{{40\sqrt x }}\\
= \dfrac{{\sqrt x - 5}}{{20}}\\
Do:x \ge 0\\
\to \sqrt x \ge 0\\
\to \sqrt x - 5 \ge - 5\\
\to \dfrac{{\sqrt x - 5}}{{20}} \ge - \dfrac{5}{{20}}\\
\to \dfrac{{\sqrt x - 5}}{{20}} \ge - \dfrac{1}{4}\\
\to Min = - \dfrac{1}{4}\\
\Leftrightarrow x = 0
\end{array}\)
