Let $X$ be an integral scheme and let $\mathcal F$ be an $\mathcal O_X$-module of finite presentation. Show that there exists an open dense subset $U$ of $X$ and an integer $n ≥ 0$ such that $\mathcal F|_U\cong \mathcal O^n_X|_U .$
Since $\mathcal F$ is finite presentation, there exists an open set $Y=\{x\in X|\mathcal F_x\cong \mathcal O_{X,x}^n\}$ for $n\ge 0$. And $X$ is irreducible, the $n$ is unique. Consider the kernel $K$ of the map $\mathcal O^n_X|_U\rightarrow\mathcal F|_U\rightarrow0.$ If take an affine open neighborhood Spec$A$$\subset U$ of $x$, then for all $p_x\in$ Spec $A$, $K_{p_x}=0$?
Qestion: Is this correct? How can I show that there is an open set $U$ s.t. $\mathcal O^n_X|_U\cong\mathcal F|_U.$
Question: "How can I show that there is an open set U s.t. OnX|U≅F|U."
Answer: If you look in Matsumuras book you will find the following result: If $A$ is a noetherian integral domain and $M$ a finitely generated $A$-module, there is an element $0 \neq a\in A$ with $M_a$ a free $A_a$-module. If $\tilde{M}$ is the sheaf of $\mathcal{O}$-modules on $X:=Spec(A)$, this result implies that $\tilde{M}_{D(a)} \cong \mathcal{O}_{D(a)}^n$ for some $n$.
In your case, choose any open affine subscheme $U:=Spec(A) \subseteq X$. Since $X$ is integral it follows $U$ is dense. Let $F(U):=M$. It follows $D(a) \subseteq U$ is an open dense subscheme with this property: There is an isomorphism
$$F_{D(a)} \cong (\mathcal{O}_{D(a)})^n.$$