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