Đáp án:
$a)A=\sqrt{x}-1\\ b) x=n^2(n \in \mathbb{N}, n>1)$
Giải thích các bước giải:
$A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\\ ĐKXĐ: \left\{\begin{array}{l} x \ge 0\\ \sqrt{x}-1 \ne 0 \\x-\sqrt{x} \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\ \sqrt{x} \ne 1 \\\sqrt{x}(\sqrt{x}-1) \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x > 0\\ x \ne 1 \end{array} \right.\\ a)A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\\ =\dfrac{x}{\sqrt{x}-1}-\dfrac{\sqrt{x}(2\sqrt{x}-1)}{(\sqrt{x}(\sqrt{x}-1)}\\ =\dfrac{x}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\\ =\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}\\ =\dfrac{(\sqrt{x}-1)^2}{\sqrt{x}-1}\\ =\sqrt{x}-1\\ b)A \in \mathbb{Z}\\ \Leftrightarrow \sqrt{x}-1 \in \mathbb{Z}\\ \Leftrightarrow \sqrt{x} \in \mathbb{Z}\\ \Rightarrow \left[\begin{array}{l} x=0\\x=1\\x=n^2(n \in \mathbb{N}, n>1)\end{array} \right.\\ \text{Kết hợp điều kiện}\\ \Rightarrow x=n^2(n \in \mathbb{N}, n>1)$
