I've been asked this question by a colleague who's not an algebraic geometer; we both feel that the answer should be "no", but I don't have a clue how to prove it. Here's the question: let $X$ be a smooth rational variety (over the complex numbers, say). Is it true that every point of $X$ has a Zariski open neighbourhood that is isomorphic to an open subset of ${\mathbb P}^n$?
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Some partial results related to this open problem can be found in the very recent preprint by F. Bogomolov and C. Böhning On uniformly rational varieties, see arXiv:1307.0102.
According to the authors (see the Introduction) this question was first raised by M. Gromov in his paper
Oka's principle for holomorphic sections of elliptic bundles, Journal of the American Mathematical Society 2, Vol. 4 (1989).

Francesco Polizzi
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Here is a preprint of Ilya Karzhemanov constructing counterexamples in dimensions $n\geq 4$:

John Pardon
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By the way, I would have thought that the answer to your question (not the stronger one) should be "yes", but as many people think the converse, I am now confused. If a point admits no such neighbourhood, it implies that every birational maps $X\to \mathbb{P}^n$ is either not defined at $x$ or contracts something through $x$. Do you have some candidate for $x$ and $X$?
– Jérémy Blanc Oct 10 '12 at 21:08