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