Đáp án:
$\min A = -1\Leftrightarrow x = 0$
Giải thích các bước giải:
$+)\quad A = \dfrac{\sqrt x -5}{\sqrt x +5}$
$\to A = 1 -\dfrac{10}{\sqrt x +5}$
$A\in \Bbb Z \to \dfrac{10}{\sqrt x + 5}\in\Bbb Z$
$\to \sqrt x + 5 \in Ư(10)=\{-10;-5;-2;-1;1;2;5;10\}$
Ta có:
$\sqrt x \geq 0$
$\to \sqrt x + 5 \geq 5$
Do đó:
$\sqrt x + 5 =\{5;10\}$
$\to \sqrt x =\{0;5\}$
$\to x = \{0;25\}$
$+)\quad A = 1 -\dfrac{10}{\sqrt x +5}$
Ta có:
$\sqrt x + 5 \geq 5$
$\to \dfrac{10}{\sqrt x +5}\leq 2$
$\to -\dfrac{10}{\sqrt x +5}\geq -2$
$\to 1 -\dfrac{10}{\sqrt x +5}\geq -1$
$\to A \geq -1$
Dấu $=$ xảy ra $\Leftrightarrow \sqrt x = 0\Leftrightarrow x = 0$
Vậy $\min A = -1\Leftrightarrow x = 0$
