Đáp án:
1, A < 1
2, A > N
3, x = 25 ; x = 4
Giải thích các bước giải:
ĐKXĐ : x $\neq$ 1 ; x ≥ 0
1, A - 1 = $\dfrac{ \sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{-2}{\sqrt{x}+1}<0$ ∀x∈ĐKXĐ
2, A - N = $\dfrac{2x-2\sqrt{x}-x+2\sqrt{x}+3}{2x+2\sqrt{x}}=\dfrac{x+3}{2x+2\sqrt{x}}>0$ ∀x∈ĐKXĐ
3, $\dfrac{2}{A}=\dfrac{2\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{2\sqrt{x}-2+4}{\sqrt{x}-1}=2+\dfrac{4}{\sqrt{x}-1}$
$\dfrac{2}{A}$ nguyên ⇔ 4 chia hết cho $\sqrt{x}-1$
⇔ \(\left[ \begin{array}{l}\sqrt{x}-1=4\\\sqrt{x}-1=1\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=25\\x=4\end{array} \right.\)
