a. DKXD: \(\left\{{}\begin{matrix}x^2-2x\ne0\\x-2\ne0\\x+2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne2\\x\ne-2\end{matrix}\right.\)
\(A=\left(\dfrac{2}{x^2-2x}+\dfrac{1}{x-2}\right):\dfrac{x+2}{x^2-4x+4}\)
\(=\left(\dfrac{2}{x\left(x-2\right)}+\dfrac{1}{x-2}\right):\dfrac{x+2}{\left(x-2\right)^2}\)
\(=\dfrac{2+x}{x\left(x-2\right)}.\dfrac{\left(x-2\right)^2}{x+2}\)\(=\dfrac{x-2}{x}\)
b. Khi \(x=\dfrac{1}{9}\Rightarrow A=\dfrac{\dfrac{1}{9}-2}{\dfrac{1}{9}}=-17\)
c. Để A nguyên \(\Leftrightarrow\dfrac{x-2}{x}\) nguyên \(\Leftrightarrow1-\dfrac{2}{x}\) nguyên \(\Leftrightarrow2⋮x\)\(\Leftrightarrow x\inƯ\left(2\right)\)
\(\Leftrightarrow x\in\left\{1;-1;2;-2\right\}\)
Kết hợp DKXD \(\Rightarrow x\in\left\{1;-1\right\}\)
\(\left(a\right):A=\left(\dfrac{2}{x^2-2x}+\dfrac{1}{x-2}\right):\dfrac{x+2}{x^2-4x+4}\left(x\ne\left\{0;\pm2\right\}\right)\\ =\left(\dfrac{2}{x\left(x-2\right)}+\dfrac{1}{x-2}\right):\dfrac{x+2}{\left(x-2\right)^2}\\ =\dfrac{2+x}{x\left(x-2\right)}.\dfrac{\left(x-2\right)^2}{x+2}=\dfrac{x-2}{x}\)
\(\left(b\right):x=\dfrac{1}{9}\left(TMDK\right)=>A=\dfrac{\dfrac{1}{9}-2}{\dfrac{1}{9}}=-\dfrac{17}{9}:\dfrac{1}{9}=-\dfrac{17}{9}.9=-17\)
\(\left(c\right):A=\dfrac{x-2}{x}=\dfrac{x}{x}-\dfrac{2}{x}=1-\dfrac{2}{x}\\ A\in Z=>\dfrac{2}{x}\in Z\\ =>2⋮x=>x\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\\ Do:x\ne\left\{0;\pm2\right\}=>x\in\left\{\pm1\right\}\)
\(Vậy\ các\ giá\ trị\ nguyên\ của\ x\ thỏa\ mãn\ A\ nguyên\ là :1;-1\)
