Why $|x|$ is not rational expression?

If there were polynomials $p$ and $q$ such that $|x|=\frac{p(x)}{q(x)}$ for all $x\in\Bbb R$, then $\frac{p(x)}{q(x)}$ would be defined for all $x\in\Bbb R$, and $q(x)$ could never be $0$. Moreover, we’d have $p(x)=|x|q(x)$ for all $x\in\Bbb R$. If $x\ge 0$, this means that $p(x)=xq(x)$, and if $x<0$ it means that $p(x)=-xq(x)$.

Let $r(x)=p(x)-xq(x)$; this is certainly a polynomial, and

$$r(x)=\begin{cases} 0,&\text{if }x\ge 0\\ -2xq(x),&\text{if }x<0\;. \end{cases}$$

I’m going to assume that you know that a polynomial of degree $n\ge 1$ has at most $n$ real roots, though you’ve probably never seen a proof. Our supposed polynomial $r(x)$ evidently has infinitely many real roots, since each non-negative real is a root, so it must be the constant function $r(x)\equiv 0$. But then $-2xq(x)=0$ for each $x<0$, and it follows that $q(x)=0$ for each $x<0$. We saw at the beginning that this is impossible: $q(x)$ can never be $0$. This contradiction shows that in fact no such polynomials $p$ and $q$ exist, and $|x|$ is not a rational function.

If the value of a polynomial function is specified at infinitely many points, then the polynomial is determined completely. So, if we just look at $x \geq 0$ then we must have $p(x) = x q(x)$. Similarly, we must have $p(x) = -x q(x)$. But $x q(x) \neq -x q(x)$ unless $q(x) = 0$.

For instance, because a rational function has a derivative at each point of its domain, and $\lvert x\rvert$ doesn't have one at $0$.