Đáp án:
$a)\text{ĐKXĐ:} \left\{\begin{array}{l} a > 0\\a \ne 1 \end{array} \right.\\ A=\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ b) a=4\\ c)x \in \varnothing$
Giải thích các bước giải:
$A=\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\\ a)\text{ĐKXĐ:}\left\{\begin{array}{l} a \ge 0\\a-\sqrt{a} \ne 0 \\ \sqrt{a}-1 \ne 0 \\ a-2\sqrt{a}+1 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} a \ge 0\\\sqrt{a} (\sqrt{a}-1)\ne 0 \\ \sqrt{a} \ne 1 \\ (\sqrt{a}-1)^2 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} a \ge 0\\ a \ne 0 \\ a \ne 1 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} a > 0\\a \ne 1 \end{array} \right.\\ A=\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\\ =\left(\dfrac{1}{\sqrt{a}(\sqrt{a}-1)}+\dfrac{1}{\sqrt{a}-1}\right).\dfrac{a-2\sqrt{a}+1}{\sqrt{a}+1}\\ =\left(\dfrac{1}{\sqrt{a}(\sqrt{a}-1)}+\dfrac{\sqrt{a}}{\sqrt{a}(\sqrt{a}-1)}\right).\dfrac{(\sqrt{a}-1)^2}{\sqrt{a}+1}\\ =\dfrac{\sqrt{a}+1}{\sqrt{a}(\sqrt{a}-1)}.\dfrac{(\sqrt{a}-1)^2}{\sqrt{a}+1}\\ =\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ b)A=\dfrac{1}{2}\\ \Leftrightarrow \dfrac{\sqrt{a}-1}{\sqrt{a}}=\dfrac{1}{2}\\ \Leftrightarrow \dfrac{\sqrt{a}-1}{\sqrt{a}}-\dfrac{1}{2}=0\\ \Leftrightarrow \dfrac{2(\sqrt{a}-1)-\sqrt{a}}{2\sqrt{a}}=0\\ \Leftrightarrow \dfrac{\sqrt{a}-2}{2\sqrt{a}}=0\\ \Leftrightarrow \sqrt{a}-2=0\\ \Leftrightarrow a=4\\ c)A \in \mathbb{Z}\\ \Leftrightarrow \dfrac{\sqrt{a}-1}{\sqrt{a}} \in \mathbb{Z}\\ \Leftrightarrow 1 - \dfrac{1}{\sqrt{a}} \in \mathbb{Z}\\ \Rightarrow \dfrac{1}{\sqrt{a}} \in \mathbb{Z}\\ a \in \mathbb{Z} \Rightarrow \sqrt{a} \in Ư(1)\\ \Leftrightarrow \sqrt{a} \in \{\pm 1\}\\ \Rightarrow a=1(L)$
