Đặt \(A=\frac{3x^2-4x-17}{x+2}\)
\(A=\frac{3x^2+6x-10x-20+3}{x+2}=\frac{3x.\left(x+2\right)-10.\left(x+2\right)+3}{x+2}=\frac{\left(x+2\right).\left(3x-10\right)+3}{x+2}\)
\(A=\frac{\left(x+2\right).\left(x-10\right)}{x+2}+\frac{3}{x+2}=x-10+\frac{3}{x+2}\)
Do x nguyên => x - 10 nguyên
Để A nguyên thì \(\frac{3}{x+2}\) nguyên
\(\Rightarrow3⋮x+2\)
\(\Rightarrow x+2\in\left\{1;-1;3;-3\right\}\)
\(\Rightarrow x\in\left\{-1;-3;1;-5\right\}\)
Vậy \(x\in\left\{-1;-3;1;-5\right\}\) thì \(\frac{3x^2-4x-17}{x+2}\) nguyên