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