Suppose it asks to show $\mathbb{Z}[x]/(2x-1) \cong \mathbb{Z}[\frac{1}{2}]$
Cand I do like this ? First of all elements of $\mathbb{Z}[x]/(2x-1)$ is of form $\frac{m}{2^n}+(2x-1)$ and also all elements of $\mathbb{Z}[\frac{1}{2}]$ are of form $\frac{m}{2^n}$. Now define a normal projection $\mathbb{Z}[\frac{1}{2}] \to \mathbb{Z}[x]/(2x-1)$ where $a \to a+(2x-1)$ for all $a\in \mathbb{Z}[\frac{1}{2}]$. So kernal is srt of all $a$ such that $2x-1|a \implies a=0$ so both are isomorphic.
On
In almost every introduction to algebra you will learn that quotient rings (quotient groups, quotient spaces etc.) consist of cosets and that it is good to work with cosets. As far as I can tell, this is quite wrong. In order to work with and understand quotient rings, one really has to learn them in a different way.
If $R$ is a ring and $I$ is an ideal of $R$, you should think of $R/I$ as a natural ring equipped with a surjective homomorphism $p : R \to R/I$ with $\ker(p)=I$. This means in particular that $p(a)=p(b)$ iff $a-b \in I$. Once $R/I$ with this property is constructed, you may forget about cosets completely. You will never need them again. Far more important is the fundamental theorem on homomorphisms: Any homomorphism $f : R \to S$ with $f|_I = 0$ extends uniquely to a homomorphism $\overline{f} : R/I \to S$. This means that $R/I$ is the "universal ring over $R$ which kills $I$".
In particular, homomorphisms $\mathbb{Z}[x]/(2x-1) \to S$ correspond to homomorphisms $f : \mathbb{Z}[x] \to S$ with $f(2x-1)=0$, i.e. $2 f(x)=1$. Since $f$ is completely determined by $f(x)$, we see that these homomorphisms correspond to elements $s \in S$ such that $2s=1$. One should think of $\mathbb{Z}[x]/(2x-1)$ as the universal ring which has a solution to the equation $2x=1$.
Now, this makes it quite obvious how to construct a homomorphism $\mathbb{Z}[x]/(2x-1) \to \mathbb{Z}[\frac{1}{2}]$, namely it maps $x \mapsto \frac{1}{2}$. This is well-defined and unique by the discussion above.
Another misconception is that isomorphisms are homomorphisms which happen to be injective and surjective. It is better and more natural to think of isomorphisms as invertible homomorphisms. So let's construct an inverse to our given homomorphism.
Specifically, we define $\mathbb{Z}[\frac{1}{2}] \to \mathbb{Z}[x]/(2x-1)$ by $\frac{a}{2^k} \mapsto a x^k$. Now one has to check that this is well-defined and a ring homomorphism. For well-definedness, notice that $\frac{a}{2^k}=\frac{b}{2^n}$ means $2^n a = 2^k b$, which implies $a x^k = b x^n$ after multiplying with $x^k x^n$. After that, one easily checks that it is inverse to the given one. In fact, it suffices to check this on the generators, where it is trivial.
More generally, if $A$ is any commutative ring and $f \in A$, then $A[\frac{1}{f}] \cong A[x]/(fx-1)$. The abstract reason is simply: Both rings have the same universal property, they are the "smallest" ring extensions of $A$ which make $f$ invertible. Hence, they are isomorphic.
An idea: define
$$\phi: \Bbb Z[x]\to\Bbb Z\left[\frac12\right]\;,\;\;\phi(f(x)):=f\left(\frac12\right)$$
Check the above is an epimorphism and in fact
$$\ker\phi=\left\{\;f(x)\in\Bbb Z[x]\;:\;\;f\left(\frac12\right)=0\;\right\}=\langle 2x-1\rangle$$
Edited on request:
We can do as follows: let $\;0\neq f(x)\in\ker\phi\;$ . If $\;f(x)=ax+b\;$ , then
$$f\left(\frac12\right)=\frac a2+b=0\implies a=-2b\implies f(x)=-2bx+b=-b(2x-1)\in\langle 2x-1\rangle$$
Now induction of $\;\deg f\;$ . First, we can work within the rational polynomials as $\;\Bbb Z[x]\subset \Bbb Q[x]\;$ , so
$$f\left(\frac12\right)=0\iff f(x)=\left(x-\frac12\right)q(x)\;,\;\;q(x)\in\Bbb Q[x]$$
use now Gauss's Lemma: since we have an integer polynomial reducible over the rationals, this same polynomial must be reducible over the integers, so
$$\exists\,h(x)\,,\,g(x)\in\Bbb Z[x]\;\;s.t.\;\;f(x)=h(x)g(x)$$
But then
$$0=f\left(\frac12\right)=h\left(\frac12\right)g\left(\frac12\right)\left(\frac12\right)\implies h\left(\frac12\right)=0\;\;or\;\;g\left(\frac12\right) $$
as we're in an integer domain, and since $\;\deg h\,,\,\deg g<\deg f\;$ we can apply the inductive hypothesis to get that $\;f(x)\;$ must be a multiple (in $\;\Bbb Z[x]\;$ !) of $\;2x-1\;$