, 2008

First note that $a + 2b$ and $2a + b$ cannot both be zero, so $a$ and $b$ cannot both be zero. Eliminating the fractions in $\frac{4a}{a+2b} - \frac{5b}{2a+b} = 1$ we get $8a^2 + 4ab - 5ab - 10b^2 = 2a^2 + 5ab + 2b^2$, which we then rewrite as $6(a + b)(a - 2b) = 0$.

If $a = -b$, then $b$ must be non-zero and the value of $\frac{a-2b}{4a+5b}$ is equal to $\frac{-3b}{b} = -3$.

If, on the other hand, we have $a = 2b$, then $\frac{a-2b}{4a+5b} = 0$ (the denominator is non-zero).

The only possible values of the expression are $0$ and $-3$.