\(D=\dfrac{x^2-2x+1}{x+1}=\dfrac{\left(x+1\right)^2-4x-4+4}{x+1}=x+1-4+\dfrac{4}{x+1}=x-3+\dfrac{4}{x+1}\)
( x # -1 )
Để : D nguyên ⇔ x nguyên ⇔ x + 1∈ Ư ( 4 )
+) x + 1 = 4 ⇔ x = 3 ( TM )
+) x + 1 = - 4 ⇔ x = - 5 ( TM )
+) x + 1 = 2 ⇔ x = 1 ( TM )
+) x + 1 = - 2 ⇔ x = - 3 ( TM)
+) x + 1 = 1 ⇔ x = 0 ( TM)
+) x + 1 = - 1 ⇔ x = - 2 ( TM)
Vậy ,..........