Lời giải:
a) ĐK: \(x\neq 0;3\)
\(A=\frac{x^3-3x^2-x+3}{3x-x^2}=\frac{x^2(x-3)-(x-3)}{x(3-x)}\)
\(=\frac{(x^2-1)(x-3)}{x(3-x)}=\frac{x^2-1}{-x}=\frac{1-x^2}{x}\)
Vậy \(A=\frac{1-x^2}{x}\)
b) Ta có \(A=\frac{1-x^2}{x}=\frac{1}{x}-x\)
Với \(x\in\mathbb{Z}\) , để \(A\in\mathbb{Z}\) thì \(\frac{1}{x}\in\mathbb{Z}\)
Điều này xảy ra khi \(1\vdots x\Leftrightarrow x\in\left\{-1;1\right\}\)
Thử lại thấy thỏa mãn.
Vậy \(x=\pm 1\)
a) \(A=\dfrac{x^3-3x^2-x+3}{3x-x^2}\)
\(\Leftrightarrow A=\dfrac{x^3-3x^2-x+3}{-\left(x^2-3x\right)}\)
\(\Leftrightarrow A=\dfrac{\left(x^3-3x^2\right)-\left(x-3\right)}{-\left(x^2-3x\right)}\)
\(\Leftrightarrow A=\dfrac{x^2\left(x-3\right)-\left(x-3\right)}{-x\left(x-3\right)}\)
\(\Leftrightarrow A=\dfrac{\left(x-3\right)\left(x^2-1\right)}{-x\left(x-3\right)}\)
\(\Leftrightarrow A=\dfrac{x^2-1}{-x}\)
\(\Leftrightarrow A=\dfrac{-x^2+1}{x}\)
b) Để \(A\in Z\Leftrightarrow\dfrac{-x^2+1}{x}\in Z\Leftrightarrow-x^2+1⋮x\)
\(\Rightarrow1⋮x\)
\(\Rightarrow x\in U\left(1\right)=\left\{-1;1\right\}\)
\(\Rightarrow x\in\left\{-1;1\right\}\)
Vậy \(x=-1\) hoặc \(x=1\) thì \(A\in Z\)
a) \(A = \frac{x^{3} - 3x^{2} - x + 3}{3x - x^{2}}\) (ĐK: \(x \neq 3\))
\(= \frac{x^{2}(x - 3) - (x - 3)}{x(3 - x)}\)
\(= \frac{ -(x - 3)(x^{2}- 1)}{x(x - 3)}\)
\(= \frac{- x^{2} + 1}{x}\)
b) Ta có: A \(= \frac{- x^{2} + 1}{x}\) (ĐK: \(x \neq 3\))
\(= - x + \frac{1}{x}\)
Để A có giá trị nguyên khi: \(1\vdots x\)
Hay: \(x \epsilon\) Ư(1) = \(\left \{ -1;1 \right \}\)
Do đó:
x = - 1
x = 1
Vậy \(x \epsilon\) \(\left \{ -1;1 \right \}\) thì A có giá trị nguyên
