If $L/K$ is a field extension, then $L$ is a vector space over $K$ Although the result is intuitive, I'm still not certain about inverse elements in $L$. For instance, if $K=\mathbb Q$ and $L=\mathbb Q(\sqrt2,\sqrt3,\sqrt5,\sqrt7)$, then why is 
$$\frac{1}{\sqrt2+\sqrt3+\sqrt5+\sqrt7}$$
a linear combination of $\{1,\sqrt 2, \sqrt 3, \sqrt 6, \dots, \sqrt{210}\}$? 
Now, actually, I'm aware of the fact that in the case of extensions of $\mathbb Q$ with surds, we can compute the inverse: $$-185 \sqrt{2}+145 \sqrt{3}+133 \sqrt{5}-135 \sqrt{7}-62 \sqrt{30}+50 \sqrt{42}+34 \sqrt{70}-22\sqrt{105},$$
but this was painstaking to compute, is there an easy way to see why this holds for any field extension $L/K$?
 A: The additive law is the additive law of $L$ (and $K$) . The external law is given by multiplication in $L$: $(\lambda, x)\in K\times L\mapsto \lambda x\in L$. It is easy to check that these two laws verify the axioms of a $K$-vector space.
Note that the extension $L/K$ does not need to be algebraic (you could take $L=K(X),$ where $X$ is an inderminate).
The fact that in your example $\frac{1}{\sqrt2+\sqrt3+\sqrt5+\sqrt7}$ is 
a linear combination of $\{1,\sqrt 2, \sqrt 3, \sqrt 6, \dots, \sqrt{210}\}$ has less to do with the fact that $L$ is a $K$-vector space, but more to do with the fact that $L/K$ is algebraic, which is more a ring theoritical reason.
More precisely if $\alpha_1,\ldots,\alpha_n$ are algebraic over $K$, then $K(\alpha_1,\ldots,\alpha_n)=K[\alpha_1,\ldots,\alpha_n]$. This comes from the case $n=1$ by induction. 
A: It seems to me like you're asking two separate questions:


*

*Why is $L$ a $K$-vector space.

*How can I write a given element in terms of a given basis.


The first question has a simple answer: because $L$ satisfies all the axioms of a $K$-vector space. Indeed, if we have addition and scalar multiplication, that's it! Here, you can completely forget about the fact that $L$ is a field.
Once you know that $L$ is a $K$-vector space, you can compute its dimension to be $[L:K]$ and find a basis. Any linearly independent set of size $[L:K]$ is a basis and, armed with this information, you know abstractly that every element of $L$ can be expressed as a $K$-linear combination of this basis.
However, actually computing explicitly how a given element looks in terms of this basis is tedious, and there's no reason you should expect otherwise!
The other answers give procedures for carrying out this computation. But I want to stress: whether or not this computation is easy or hard has nothing to do with the much easier fact that $L$ is a $K$-vector space.
A: Well, I think you can find the formula iteratively.
For this, take say ${\Bbb Q}(\sqrt n) = \{a+b\sqrt n\mid a,b\in{\Bbb Q}\}$, where $n\ne0,1$ is a square-free integer.
Then the  multiplicative inverse of $a+b\sqrt n$ is given by 
$$\frac{a}{a^2-nb^2} - \frac{b}{a^2-nb^2}\sqrt n.$$
Just adjoin the square roots step by step.
First, $\frac{1}{\sqrt 2} = \frac{1}{2}\sqrt 2$.
Second, the ansatz is $\frac{1}{\sqrt 2 +\sqrt 3} = a + b\sqrt 3$, where $a,b\in{\Bbb Q}(\sqrt 2)$.
Then $1 = (a+b\sqrt 3)(\sqrt 2 + \sqrt 3)$ leads to values for $a,b$, and so on.
