Answer:
will always have an even value.
Step-by-step explanation:
We are given an integer x.
Let x be even, then it can be written in the form x = 2n, where n is an integer.
If we evaluate,
![(x^2 - x) = (2n)^2 - 2n = 4n^2 - 2n = 2(2n^2 - n)](https://tex.z-dn.net/?f=%28x%5E2%20-%20x%29%20%3D%20%282n%29%5E2%20-%202n%20%3D%204n%5E2%20-%202n%20%3D%202%282n%5E2%20-%20n%29)
Thus, it have an even value.
If we take x to be an odd integer, then,it can be written in the form x = 2n+1, where n is an integer.
![(x^2 - x) = (2n+1)^2 - 2n = 4n^2 + 2n = 2(2n^2 + n)](https://tex.z-dn.net/?f=%28x%5E2%20-%20x%29%20%3D%20%282n%2B1%29%5E2%20-%202n%20%3D%204n%5E2%20%2B%202n%20%3D%202%282n%5E2%20%2B%20n%29)
Thus, it have an even value.
Thus,
will always have an even value.