a: \(A=\dfrac{1-x}{2+x}-\dfrac{x-1}{x-2}+\dfrac{4-x^3}{4-x^2}\)
\(=\dfrac{-\left(x-1\right)}{x+2}-\dfrac{x-1}{x-2}+\dfrac{x^3-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-\left(x-1\right)\left(x-2\right)-\left(x-1\right)\left(x+2\right)+x^3-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-\left(x^2-3x+2\right)-\left(x^2+x-2\right)+x^3-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-x^2+3x-2-x^2-x+2+x^3-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^3-2x^2+2x-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2\left(x-2\right)+2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)\left(x^2+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+2}{x+2}\)
b: Thay x=-0,25 vào A, ta được:
\(A=\dfrac{\left(-0,25\right)^2+2}{-0,25+2}=\dfrac{33}{28}\)
c: Để A là số nguyên thì \(x^2+2⋮x+2\)
=>\(x^2-4+6⋮x+2\)
=>\(6⋮x+2\)
=>\(x+2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
=>\(x\in\left\{-1;-3;0;-4;1;-5;4;-8\right\}\)
