DKXD: \(\left\{{}\begin{matrix}x-1\ne0\\x\left(x-2\right)\ne0\\x+2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne0\\x\ne2\\x\ne-2\end{matrix}\right.\)
a, \(K=\left(x-3+\dfrac{1}{x-1}\right):\left(x-1-\dfrac{1}{x-1}\right):\dfrac{x+2}{x}\)
\(=\dfrac{\left(x-3\right)\left(x-1\right)+1}{x-1}:\dfrac{\left(x-1\right)^2-1}{x-1}:\dfrac{x+2}{x}\)
\(=\dfrac{x^2-4x+3+1}{x-1}:\dfrac{x^2-2x+1-1}{x-1}:\dfrac{x+2}{x}\)
\(=\dfrac{x^2-4x+4}{x-1}:\dfrac{x^2-2x}{x-1}:\dfrac{x+2}{x}\)
\(=\dfrac{\left(x-2\right)^2}{x-1}.\dfrac{x-1}{x\left(x-2\right)}.\dfrac{x}{x+2}\)
\(=\dfrac{\left(x-2\right)^2\left(x-1\right)x}{\left(x-1\right)x\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x-2}{x+2}\)
b, Để K nguyên \(\Leftrightarrow\dfrac{x-2}{x+2}\) nguyên \(\Leftrightarrow\dfrac{x+2-4}{x+2}\) nguyên
\(\Leftrightarrow1-\dfrac{4}{x+2}\)nguyên \(\Leftrightarrow\dfrac{4}{x+2}\) nguyên \(\Leftrightarrow4⋮\left(x+2\right)\)\(\Leftrightarrow\left(x+2\right)\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
\(\Leftrightarrow x\in\left\{-1;-3;0;-4;2;-6\right\}\)
Kết hợp DKXD \(\Rightarrow x\in\left\{-1;-3;-4;-6\right\}\)
a: \(K=\dfrac{x^2-4x+3+1}{\left(x-1\right)}:\dfrac{x^2-2x+1-1}{x-1}:\dfrac{x+2}{x}\)
\(=\dfrac{\left(x-2\right)^2}{x-1}\cdot\dfrac{x-1}{x\left(x-2\right)}\cdot\dfrac{x}{x+2}\)
\(=\dfrac{\left(x-2\right)}{x+2}\)
b: Để K là số nguyên thì x+2-4 chia hết cho x+2
\(\Leftrightarrow x+2\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{-1;-3;-4;-6\right\}\)
