I feel like this question is probably easy, but I'm just missing something obvious. I usually pick things up quickly (especially in algebra) but algebraic geometry continues to give me fits.

We will work over a fixed algebraically closed field k.

If X is an affine or quasi-affine variety, denote by \mathcal{O}(X) the ring of regular functions on X (i.e., f\in\mathcal{O}(X)\Leftrightarrow \forall P\in X,\exists an open neighborhood U_P of P such that f|_U is given by a rational function \frac{g(x,y)}{h(x,y)}\in k(x,y).

There are two questions here:

1. Show that \mathcal{O}(\mathbb{A}^2\setminus\mathcal{Z}(x))=k[x,y][x^{-1}] (where \mathcal{Z}(x) denotes the zero set of x which, over \mathbb{A}^2, is simply \mathcal{Z}(x)=\{(0,y)|y\in k\}).

2. Show that \mathcal{O}(\mathbb{A}^2\setminus\{(0,0)\})=\mathc  al{O}(\mathbb{A}^2). (It is already known that \mathcal{O}(\mathbb{A}^2)=k[x,y]).

I really feel like I'm just missing something blatantly obvious. I can intuitively believe #1: if we only consider those points where x is nonzero, then \frac{1}{x} is defined everywhere, and obviously a rational function. But to actually PROVE that the given ring is exactly equal to the ring of regular functions on the set where x\neq 0 is a tougher matter... I think...

Any help would be appreciated. I really, really like this subject, but I'm having so much trouble understanding it fully, or at least to the point where I can competently work with it...