In this exercise it is asked to compute the ring of fractions of a noetherian ring $R$ defined as $$\varinjlim_{U\in \mathfrak{U}}\mathcal{O}_X(U)$$ where $X=\text{Spec}\ R$ and $\mathfrak{U}$ is the set of dense open subsets of $X$.
I believe the setting is too general to be able to describe this ring as usual. $R$ should be supposed to be also reduced to get it. Does anybody have a proof in the general setting of $R$ only supposed to be noetherian?
Edit: the exercise actually asks not for the usual total ring of fractions, but for the ring of rational fractions, as it is mentioned in the answer (but I did not know the definition of such a ring)
Lemma. You can replace $\mathfrak{U}$ by the set $\mathfrak{U}'$ of elementary affine open subsets $D(f)$ such that $f$ does not belong to any minimal prime of $R$ (which is equivalent to say that $D(f)$ is dense in $X$).
Proof. Take $U\in\mathfrak{U}$. Then you can write $F = X\backslash U = V(I)$ for some ideal $I$ of $R$. As $U$ is dense, $F$ does not contain any irreducible component of $X$ (if $U$ doesn't cut an irreducible component then it is contained in the union of the finitely number of others irreducible components, then the closure of $U$ is also be contained in the latter union and $U$ is not dense) and this translates algebraically into the fact that the ideal $I$ is not contained in any minimal prime ideal of $R$. Now if $\mathfrak{p}_x\in U$ you can find (by prime avoidance) an $f\in I$ such that $\mathfrak{p}_x \in D(f)$ and such that $f$ doesn't belong to any minimal prime of $R$, which translates topologically into the fact that $D(f)\subset U$ is dense in $X$. So any element of $\mathfrak{U}$ is covered by sets of $\mathfrak{U}'$, which concludes the lemma.
From this you see that the direct limit you want to calculate is the localization of $R$ at the multiplicative set of elements not contained in any minimal prime. If $R$ is a domain as first supposed in the exercice, then you land in the total ring of fractions of $R$.
What about the case where $R$ is noetherian (but not a domain) ? You land on the ring of rational functions of $X$. But it's not that easy as the formulation of the exercise can let think. (See EGA I, proposition 7.1.5.)