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Since n and m are consecutive integers, the difference will always be either 1 or -1.
E.g.: n=5, m=6

\(\frac{(5-6)^x}{(6-5)^x}\)= \(\frac{(-1)^x}{(1)^x}\)

And since we have the even exponent the resolved exponent will always be 1. Thus result 1.
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Bunuel
If x is a positive even integer, and n and m are consecutive integers, then (n - m)^x/(m - n)^x =

Lets plug in some values -

x = 2
m = 4 ; n = 5

\(\frac{(n - m)^x}{(m - n)^x}\) = \(\frac{(5 - 4)^2}{(4 - 5)^2}\) =\(\frac{1^2}{-1^2}\) = 1

Hence among the given options IMHO answer is D. 1
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X is even
N and M are consecutive, thus their difference will always be -1 or 1.
Anything raised to even power is positive and 1 to power anything is 1
Thus OME : D
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Bunuel
If x is a positive even integer, and n and m are consecutive integers, then (n - m)^x/(m - n)^x =

A. -2
B. -1
C. 0
D. 1
E. 2

Notice that (n - m)^x / (m - n)^x = [(n - m)/(m - n)]^x = (-1)^x

Since x is a positive even integer, (-1)^x will always be equal to 1.

Alternate solution:

Let’s use some strategic numbers for n, m, and x.

n = 4, m = 3, and x = 2, thus:

[(4 - 3)^2]/(3 - 4)^2 = 1^2/(-1)^2 = 1

Answer: D
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Bunuel
If x is a positive even integer, and n and m are consecutive integers, then (n - m)^x/(m - n)^x =

A. -2
B. -1
C. 0
D. 1
E. 2

If power is even, then base can never be negative. Therefore, A & B are out.

Try n = 1, m = 2, x = 2

\(\frac{(1 - 2)^2}{(2 - 1)^2}\)

\(\frac{(-1)^2}{(1)^2}\)

1

(D)
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