Question

Answer 1

a: ĐKXĐ: \(x\notin\left\{3;1;-1\right\}\)

Sửa đề: \(A=\dfrac{2\left(x^2-2x-3\right)}{x^3-3x^2-x+3}\)

\(=\dfrac{2\left(x-3\right)\left(x+1\right)}{x^2\left(x-3\right)-\left(x-3\right)}\)

\(=\dfrac{2\left(x-3\right)\left(x+1\right)}{\left(x-3\right)\left(x^2-1\right)}\)

\(=\dfrac{2\left(x+1\right)}{x^2-1}=\dfrac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x-1}\)

b: Thay x=-2 vào A, ta được:

\(A=\dfrac{2}{-2-1}=\dfrac{2}{-3}=\dfrac{-2}{3}\)

c: Để A là số nguyên thì \(x-1\inƯ\left(2\right)\)

=>\(x-1\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{2;0;3;-1\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{2;0\right\}\)