Đáp án:
Min=8
Giải thích các bước giải:
\(\begin{array}{l}
P = \dfrac{{2x}}{{\sqrt x - 1}} = \dfrac{{2\left( {x - 1} \right) + 2}}{{\sqrt x - 1}}\\
= \dfrac{{2\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) + 2}}{{\sqrt x - 1}}\\
= 2\left( {\sqrt x + 1} \right) + \dfrac{2}{{\sqrt x - 1}}\\
= 2\left( {\sqrt x - 1} \right) + \dfrac{2}{{\sqrt x - 1}} + 4\\
Do:x > 1\\
BDT:2\left( {\sqrt x - 1} \right) + \dfrac{2}{{\sqrt x - 1}} \ge 2\sqrt {2\left( {\sqrt x - 1} \right).\dfrac{2}{{\sqrt x - 1}}} \\
\to 2\left( {\sqrt x - 1} \right) + \dfrac{2}{{\sqrt x - 1}} \ge 2.2\\
\to 2\left( {\sqrt x - 1} \right) + \dfrac{2}{{\sqrt x - 1}} + 4 \ge 8\\
\to Min = 8\\
\Leftrightarrow 2\left( {\sqrt x - 1} \right) = \dfrac{2}{{\sqrt x - 1}}\\
\to {\left( {\sqrt x - 1} \right)^2} = 1\\
\to \sqrt x - 1 = 1\\
\to x = 4
\end{array}\)
