Đáp án:
$\min A =\dfrac34 \Leftrightarrow x = 16$
Giải thích các bước giải:
$A = \dfrac{x - 2\sqrt x + 4}{x}\qquad (x > 0)$
$\to A = 1 - \dfrac{2}{\sqrt x} + \dfrac4x$
$\to A = \dfrac{3}{4} +\dfrac14 - 2\cdot\dfrac12\cdot\dfrac{2}{\sqrt x} +\left(\dfrac{2}{\sqrt x}\right)^2$
$\to A = \left(\dfrac{2}{\sqrt x} -\dfrac12\right)^2 +\dfrac34$
Ta có:
$\left(\dfrac{2}{\sqrt x} -\dfrac12\right)^2 \geq 0\quad \forall x > 0$
$\to \left(\dfrac{2}{\sqrt x} -\dfrac12\right)^2 +\dfrac34 \geq \dfrac34$
$\to A \geq \dfrac34$
Dấu $=$ xảy ra $\Leftrightarrow \dfrac{2}{\sqrt x} =\dfrac12 \Leftrightarrow x = 16$
Vậy $\min A =\dfrac34 \Leftrightarrow x = 16$
