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