a)\(P=\frac{2x^2-8x+8}{x^3-6x^2+12x-8}\left(x\ne2\right)\)
\(P=\frac{2\left(x-2\right)^2}{\left(x-2\right)^3}\)
\(P=\frac{2}{x-2}\)
b)Để P nguyên thì \(2⋮x-2\).Hay \(\left(x-2\right)\inƯ\left(2\right)\)
Ư(2) là:[1,-1,2,-2]
Do đó ta có bảng sau:
x-2 | -2 | -1 | 1 | 2 |
x | 0 | 1 | 3 | 4 |