Simplify the expression rac(x+1)(4a2-ax)x2-4ax-4a+x and find two consecutive even numbers whose sum is 270.

Factor out $a$ from the second term in the numerator:

Group terms in the denominator and factor by grouping:

Substitute the factored forms back into the expression:

Notice that $(4a-x) = -(x-4a)$.

Assuming $x \ne -1$ and $x \ne 4a$, we can cancel out the common factors $(x+1)$ and $(x-4a)$:

The simplified expression is $\boxed{-a}$.

Let the first even number be $x$.
The next consecutive even number is $x+2$.

The first number is $x = 134$.
The second number is $x+2 = 134+2 = 136$.
The two consecutive even numbers are $\boxed{134 \text{ and } 136}$.

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