50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010)

Only $x=0$ and $x=1$.

Obviously $x\in [0,1]$ and $0\le (\sqrt{x}{)}^{2009}+(\sqrt{1-x}{)}^{2010}<1$. On the other hand we have $x\le 1$ and $1-x\le 1$ which implies $(\sqrt{x}{)}^{2009}+(\sqrt{1-x}{)}^{2010}\le x<(1-x)=1$. The case $(\sqrt{x}{)}^{2009}+(\sqrt{1-x}{)}^{2010}=0$ is impossible, so we have to consider only the case $(\sqrt{x}{)}^{2009}+(\sqrt{1-x}{)}^{2010}=1$. Equality in the previous inequality obtains iff $(\sqrt{x}{)}^{2009}=x$ and $(\sqrt{1-x}{)}^{2010}=1-x$. This is possible only when $x=0$ or $x=1$. It's easy to see that both values satisfy the statement.