Đáp án:
b. Min=3
Giải thích các bước giải:
\(\begin{array}{l}
a.B = \left[ {\dfrac{{\sqrt x + 1 + x}}{{\sqrt x \left( {\sqrt x + 1} \right)}}} \right]:\dfrac{{\sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x }}\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x }} = \sqrt x + 1 + \dfrac{1}{{\sqrt x }}\\
b.B = \sqrt x + 1 + \dfrac{1}{{\sqrt x }}\\
Do:x > 0\\
\to BDT:Cô - si:\sqrt x + \dfrac{1}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{1}{{\sqrt x }}} \\
\to \sqrt x + \dfrac{1}{{\sqrt x }} \ge 2\\
\to \sqrt x + \dfrac{1}{{\sqrt x }} + 1 \ge 3\\
\to Min = 3\\
\Leftrightarrow \sqrt x = \dfrac{1}{{\sqrt x }}\\
\Leftrightarrow x = 1
\end{array}\)
