Slovenija 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 $-3$.