Giải thích các bước giải:
a.Ta có:
$A=\dfrac{x^2+1}{x}$
$\to A-\dfrac52=\dfrac{x^2+1}{x}-\dfrac52$
$\to A-\dfrac52=\dfrac{2(x^2+1)-5x}{x}$
$\to A-\dfrac52=\dfrac{2x^2-5x+2}{x}$
$\to A-\dfrac52=\dfrac{(x-2)(2x-1)}{x}$
Mà $0<x\le \dfrac12\to x-2<0, 2x-1\le 0$
$\to \dfrac{(x-2)(2x-1)}{x}\ge 0$
$\to A-\dfrac52\ge 0$
$\to A\ge\dfrac52$
Dấu = xảy ra khi $x=\dfrac12$
b.Ta có:
$A-\dfrac52=\dfrac{(x-2)(2x-1)}{x}$
Do $x\ge 2\to x-2\ge 0, 2x-1>0$
$\to \dfrac{(x-2)(2x-1)}{x}\ge 0$
$\to A-\dfrac52\ge 0$
$\to A\ge\dfrac52$
Dấu = xảy ra khi $x=2$
