How to add two compound fractions with fractions in numerator like this one: 1x2+23xx or...

Let us start with the properties:
Division by a number or fraction is the same as multiplication by its inverse or reciprocal.
Division by r is equal to the multiplication by ${\displaystyle \frac{1}{r}}$
${\displaystyle \frac{\frac{p}{q}}{r}=\frac{p}{q}\cdot \frac{1}{r}=\frac{p\cdot 1}{q\cdot r}=\frac{p}{qr},\left(1\right)}$
Division by ${\displaystyle \frac{t}{u}}$ is equal to the multiplication by $\frac{u}{t}:$
${\displaystyle \frac{s}{\frac{t}{u}}=s\cdot \frac{u}{t}=\frac{s\cdot u}{t}=\frac{su}{t}.\left(2\right)}$
Sum of fractions
${\displaystyle \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}.\left(3\right)}$
Apply (1) to
${\displaystyle \frac{\frac{1}{x}}{2}=\frac{1}{x}\cdot \frac{1}{2}=\frac{1\cdot 1}{x\cdot 2}=\frac{1}{2x}}$
and (2) to
${\displaystyle \frac{\frac{2}{3x}}{x}=\frac{2}{3x}\cdot \frac{1}{x}=\frac{2\cdot 1}{3x\cdot x}=\frac{2}{3x^2}}$
So by (3) we have
${\displaystyle \frac{\frac{1}{x}}{2}+\frac{\frac{2}{3x}}{x}=\frac{1}{2x}+\frac{2}{3x^2}=\frac{1\cdot 3x^2+2\times 2x}{2x\cdot 3x^2}=\frac{3x^2+4x}{6x^3}}$
${\displaystyle =\frac{x(3x+4)}{x\left(6x^2\right)}=\frac{3x+4}{6x^2}}$
For
$\frac{x}{\frac{2}{x}}+\frac{\frac{1}{x}}{x}$
We have
${\displaystyle \frac{x}{\frac{2}{x}}+\frac{\frac{1}{x}}{x}=\frac{1}{x\cdot x}=\frac{x^2}{2}+\frac{1}{x^2}=\frac{x^2\cdot x^2+2\cdot 1}{2\cdot x^2}=\frac{x^4+2}{2x^2}}$
We can apply the property Division by a fraction is the same as multiplication by its inverse or reciprocal to the following fraction