Đáp án:
2) \(M = \dfrac{{x + 7}}{{\sqrt x + 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)Thay:x = 16\\
\to B = \dfrac{1}{{\sqrt {16} - 3}} = 1\\
2)A = \dfrac{{x + \sqrt x + 10 - \sqrt x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x + 7}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
M = A:B = \dfrac{{x + 7}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}:\dfrac{1}{{\sqrt x - 3}}\\
= \dfrac{{x + 7}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\left( {\sqrt x - 3} \right)\\
= \dfrac{{x + 7}}{{\sqrt x + 3}}\\
3)M = \dfrac{{x - 9 + 16}}{{\sqrt x + 3}} = \dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + 16}}{{\sqrt x + 3}}\\
= \left( {\sqrt x - 3} \right) + \dfrac{{16}}{{\sqrt x + 3}} = \left( {\sqrt x + 3} \right) + \dfrac{{16}}{{\sqrt x + 3}} - 6\\
Do:x \ge 0\\
BDT:Co - si:\left( {\sqrt x + 3} \right) + \dfrac{{16}}{{\sqrt x + 3}} \ge 2\sqrt {\left( {\sqrt x + 3} \right).\dfrac{{16}}{{\sqrt x + 3}}} = 8\\
\to \left( {\sqrt x + 3} \right) + \dfrac{{16}}{{\sqrt x + 3}} - 6 \ge 2\\
\to Min = 2\\
\Leftrightarrow \left( {\sqrt x + 3} \right) = \dfrac{{16}}{{\sqrt x + 3}}\\
\to {\left( {\sqrt x + 3} \right)^2} = 16\\
\to \sqrt x + 3 = 4\\
\to x = 1
\end{array}\)
