Đáp án:
\(\begin{array}{l}
a)Min = - \dfrac{2}{5}\\
b)Min = 2
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0\\
P = \dfrac{{\sqrt x - 2}}{{\sqrt x + 5}} = \dfrac{{\sqrt x + 5 - 7}}{{\sqrt x + 5}}\\
= 1 - \dfrac{7}{{\sqrt x + 5}}\\
Do:\sqrt x \ge 0\forall x \ge 0\\
\to \sqrt x + 5 \ge 5\\
\to \dfrac{7}{{\sqrt x + 5}} \le \dfrac{7}{5}\\
\to - \dfrac{7}{{\sqrt x + 5}} \ge - \dfrac{7}{5}\\
\to 1 - \dfrac{7}{{\sqrt x + 5}} \ge - \dfrac{2}{5}\\
\to Min = - \dfrac{2}{5}\\
\Leftrightarrow x = 0\\
b)Q = \dfrac{{x + 5}}{{\sqrt x + 2}} = \dfrac{{x - 4 + 9}}{{\sqrt x + 2}}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) + 9}}{{\sqrt x + 2}}\\
= \left( {\sqrt x - 2} \right) + \dfrac{9}{{\sqrt x + 2}}\\
= \left( {\sqrt x + 2} \right) + \dfrac{9}{{\sqrt x + 2}} - 4\\
Do:x \ge 0\\
BDT:Co - si:\left( {\sqrt x + 2} \right) + \dfrac{9}{{\sqrt x + 2}} \ge 2\sqrt {\left( {\sqrt x + 2} \right).\dfrac{9}{{\sqrt x + 2}}} \\
\to \left( {\sqrt x + 2} \right) + \dfrac{9}{{\sqrt x + 2}} \ge 2.3\\
\to \left( {\sqrt x + 2} \right) + \dfrac{9}{{\sqrt x + 2}} \ge 6\\
\to \left( {\sqrt x + 2} \right) + \dfrac{9}{{\sqrt x + 2}} - 4 \ge 2\\
\to Min = 2\\
\Leftrightarrow \left( {\sqrt x + 2} \right) = \dfrac{9}{{\sqrt x + 2}}\\
\to {\left( {\sqrt x + 2} \right)^2} = 9\\
\to \sqrt x + 2 = 3\\
\to x = 1
\end{array}\)
