Đáp án:
a) Min=4
b) Min=2
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0\\
A = \dfrac{{x + 9}}{{\sqrt x + 4}} = \dfrac{{x - 16 + 25}}{{\sqrt x + 4}}\\
= \dfrac{{\left( {\sqrt x + 4} \right)\left( {\sqrt x - 4} \right) + 25}}{{\sqrt x + 4}}\\
= \left( {\sqrt x - 4} \right) + \dfrac{{25}}{{\sqrt x + 4}}\\
= \left( {\sqrt x + 4} \right) + \dfrac{{25}}{{\sqrt x + 4}} - 8\\
Do:x \ge 0\\
BDT:Co - si:\left( {\sqrt x + 4} \right) + \dfrac{{25}}{{\sqrt x + 4}} \ge 2\sqrt {\left( {\sqrt x + 4} \right).\dfrac{{25}}{{\sqrt x + 4}}} = 2.5\\
\to \left( {\sqrt x + 4} \right) + \dfrac{{25}}{{\sqrt x + 4}} \ge 10\\
\to \left( {\sqrt x + 4} \right) + \dfrac{{25}}{{\sqrt x + 4}} - 8 \ge 2\\
\to Min = 2\\
\Leftrightarrow \left( {\sqrt x + 4} \right) = \dfrac{{25}}{{\sqrt x + 4}}\\
\to {\left( {\sqrt x + 4} \right)^2} = 25\\
\to \sqrt x + 4 = 5\\
\to \sqrt x = 1\\
\to x = 1\\
b)B = \dfrac{{x + 24}}{{\sqrt x + 5}} = \dfrac{{x - 25 + 49}}{{\sqrt x + 5}}\\
= \dfrac{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right) + 49}}{{\sqrt x + 5}}\\
= \left( {\sqrt x - 5} \right) + \dfrac{{49}}{{\sqrt x + 5}}\\
= \left( {\sqrt x + 5} \right) + \dfrac{{49}}{{\sqrt x + 5}} - 10\\
Do:x \ge 0\\
BDT:Co - si:\left( {\sqrt x + 5} \right) + \dfrac{{49}}{{\sqrt x + 5}} \ge 2\sqrt {\left( {\sqrt x + 5} \right).\dfrac{{49}}{{\sqrt x + 5}}} \\
\to \left( {\sqrt x + 5} \right) + \dfrac{{49}}{{\sqrt x + 5}} \ge 14\\
\to \left( {\sqrt x + 5} \right) + \dfrac{{49}}{{\sqrt x + 5}} - 10 \ge 4\\
\to Min = 4\\
\Leftrightarrow \left( {\sqrt x + 5} \right) = \dfrac{{49}}{{\sqrt x + 5}}\\
\to {\left( {\sqrt x + 5} \right)^2} = 49\\
\to \sqrt x + 5 = 7\\
\to x = 4
\end{array}\)
