\(\text{a) }A=\dfrac{x-3}{x^2-6x+9}\\ A=\dfrac{x-3}{\left(x-3\right)^2}\\ \RightarrowĐKXĐ\text{ của }A:\\ \left(x-3\right)^2\ne0\\ \Leftrightarrow x-3\ne0\\ \Leftrightarrow x\ne3\)
\(\text{b) Ta có : }A=\dfrac{x-3}{\left(x-3\right)^2}\\ A=\dfrac{1}{x-3}\)
c) Mk không làm đc
\(\text{d) }\text{Để }A=2\\ thì\Rightarrow\dfrac{1}{x-3}=2\\ \Rightarrow x-3=\dfrac{1}{2}\\ \Rightarrow x=\dfrac{7}{2}\)
\(\text{e) }\text{Để }A>0\\ thì\Rightarrow\dfrac{1}{x-3}>0\\ \text{ Mà }1>0\\ \Rightarrow x-3>0\\ \Rightarrow x>3\\ \text{Để }A< 0\\ thì\Rightarrow\dfrac{1}{x-3}< 0\\ \text{ Mà }1>0\\ \Rightarrow x-3< 0\\ \Rightarrow x< 3\)
\(\text{f) }\text{Để }A\in Z\\ thì\Rightarrow1⋮x-3\\ \Rightarrow x-3\inƯ_{\left(1\right)}\\ \text{Mà }Ư_{\left(1\right)}=\left\{\pm1\right\}\)
Ta lập bảng giá trị:
| \(x-3\) | \(-1\) | \(1\) |
| \(x\) | \(2\) | \(4\) |
\(\Rightarrow x=\left\{2;4\right\}\)
Vậy............................................
