Let $K / F$ be an extension of fields. I define the two next sets: $$ F[X] = \left\{\sum_{i = 0}^n a_i X^i : n \in \mathbb{N} \mbox{, } a_i \in F \mbox{ for all } i = 0 , 1 , \ldots , n\right\} $$ and $$ F(X) = \left\{\frac{p(X)}{q(X)} : p(X) , q(X) \in F[X] \mbox{ and } q(X) \not \equiv 0\right\}\mbox{,} $$ where $X$ is a variable (it does not belong neither to $K$ nor to $F$). Now, let $x \in K$ and I define $F[x]$ as the minor ring which contains as to $F$ as to $x$, and $F(x)$ will be the minor field which contains as to $F$ as to $x$. Well now I can ask my question: are the equalities $$ F[x] = \{f(x) : f(X) \in F[X]\} \qquad \mbox{ and } \qquad F(x) = \{f(x) : f(X) \in F(X)\} $$ true? I think yes and here is my attempt: $$ F[x] \supset \{f(x) : f(X) \in F[X]\} \qquad \mbox{ and } \qquad F(x) \supset \{f(x) : f(X) \in F(X)\} $$ are clear inclusions; indeed, let $p(X) = \sum_{i = 0}^n a_i X^i$ and $q(X) = \sum_{j = 0}^m b_j X^j$ be polynomials in $F[X]$ such that $q(X) \not \equiv 0$. Since $x \in F[x]$ and it is a ring, then $$ p(x) = \sum_{i = 0}^n a_i x^i \in F[x] $$ and hence $q(x) \in F[x] \subset F(x)$ and then $\frac{1}{q(x)} \in F(x)$, assuming $q(x) \neq 0$, as it is a field. Therefore, $\frac{p(x)}{q(x)} \in F(x)$. Well, to show $$ F[x] \subset \{f(x) : f(X) \in F[X]\} \qquad \mbox{ and } \qquad F(x) \subset \{f(x) : f(X) \in F(X)\} $$ it would be enough to prove that $\{f(x) : f(X) \in F[X]\}$ is a ring and $\{f(x) : f(X) \in F(X)\}$ is a field because both sets contain as to $x$ as to $F$ (it is trivial: the polynomial $X$ belongs to $F[X]$, the constant polynomials span $F$ and $F[x] \subset F(x)$). On the other hand, they are ring and field, respectively, because $F[X]$ is a ring and $F(X)$ is a field.
Is my statement correct? Thank you by your help.
This is correct.
This isn't quite correct. What if you take $f(X) = \frac{1}{X-2}$ and $x=2$?
To fix this, you should be trying to prove that:
That is, it's not enough for $q(X)$ to be nonzero. It also needs to be nonzero when evaluated at $X=x$.
It might help you to ask yourself the question "where does the set $\{f(x) : f(X) \in F[X]\}$ live?". This set is not just an abstract set; it's a subset of $K$. Indeed, you should really define $F[x]$ as the smallest subring of $K$ containing both $F$ and $x$.
Finally, I'm not convinced by this line of your proof:
Firstly, let's deal with $F[x]$. You want to show that $\{f(x) : f(X) \in F[X]\}$ is a ring, because then it will contain $F[x]$ by definition; and your proposed method to do this works. (Putting your method more rigorously: $F[X]$ is a ring, and the map $F[X]\to K$ sending $X\mapsto x$ is a ring homomorphism. So its image must be a ring. But its image is precisely $\{f(x) : f(X) \in F[X]\}$, and this is what you wanted to show.)
You'll need to work a little harder for $F(x)$. But not much harder. You want to show that $\{p(x)/q(x) : p(X),q(X) \in F[X] \text{ and } q(x)\neq 0\}$ is a field, because then it will contain $F(x)$ by definition.
In the latter case, your method now no longer works, because $\{p(x)/q(x) : p(X),q(X) \in F[X] \text{ and } q(x)\neq 0\}$ isn't very closely related to $F(X)$ any more by what I said above. Hint: show first that it's a ring; then show that, if $p(x)/q(x) \neq 0$, it has a multiplicative inverse $(p(x)/q(x))^{-1} = q(x)/p(x) \in F(x)$.