Đáp án:
$\min f(x)= \dfrac52 \Leftrightarrow x = 3$
Giải thích các bước giải:
$\quad f(x)=\dfrac{x}{2} +\dfrac{2}{x-1}\qquad (x>1)$
$\to f(x)=\dfrac{x-1}{2} +\dfrac{2}{x-1} +\dfrac12$
$\to f(x)\geqslant 2\sqrt{\dfrac{x-1}{2}\cdot\dfrac{2}{x-1}} + \dfrac12$
$\to f(x)\geqslant 2+\dfrac12 =\dfrac52$
Dấu $=$ xảy ra $\Leftrightarrow \dfrac{x-1}{2}=\dfrac{2}{x-1}\Leftrightarrow x = 3$
Vậy $\min f(x)= \dfrac52 \Leftrightarrow x = 3$
