Question

Answer 1

a: \(H=\dfrac{x^2+9+5x-15-x^2-3x}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+2}\)

\(=\dfrac{2x-6}{x-3}\cdot\dfrac{1}{x+2}=\dfrac{2}{x+2}\)

b: Thay x=-6 vào H, ta được:

\(H=\dfrac{2}{-6+2}=\dfrac{2}{-4}=\dfrac{-1}{2}\)

c: Để H là số nguyên thì \(2⋮x+2\)

\(\Leftrightarrow x+2\in\left\{1;-1;2;-2\right\}\)

hay \(x\in\left\{-1;0;-4\right\}\)