How to add compound fractions? How to add two compound fractions with fractions in numerator like this one:
$$\frac{\ \frac{1}{x}\ }{2} + \frac{\ \frac{2}{3x}\ }{x}$$
or fractions with fractions in denominator like this one:
$$\frac{x}{\ \frac{2}{x}\ } + \frac{\ \frac{1}{x}\ }{x}$$
 A: One easy way to figure this out is that dividing by $x$ is the same as multiplying by $1/x$ (but all bets are off when $x=0$, as division by $0$ is undefined). So 
$$
\begin{align*}
\frac{ \frac{a}{b} }{c} &= \frac{1}{c} \frac{a}{b} = \frac{a}{bc} \\ \\
\frac{a}{\frac{b}{c}} &= a \frac{1}{\frac{b}{c}} = a \frac{c}{b} = \frac{ac}{b}
\end{align*}
$$
A: 
The multiplicative inverse of a fraction a/b is b/a. (Wikipedia)

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 $\dfrac{1}{r}$:
$$\dfrac{\ \dfrac{p}{q}\ }{r}=\dfrac{p}{q}\cdot \dfrac{1}{r}=\dfrac{p\cdot 1}{q\cdot r}=\dfrac{p}{q r}, \quad (1)$$
Division by $\dfrac{t}{u}$ is equal to the multiplication by $\dfrac{u}{t}$:
$$\dfrac{\ s}{\ \dfrac{t}{u}\ }=s\cdot\dfrac{u}{t}=\dfrac{s\cdot u}{t}=\dfrac{su}{t}.\quad (2)$$ 


*

*Sum of fractions
$$\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{bd}.\quad (3)$$
Apply $(1)$ to 
$$\dfrac{\ \dfrac{1}{x}\ }{2}=\dfrac{1}{x}\cdot\dfrac{1}{2}=\dfrac{1\cdot 1}{x\cdot 2}=\dfrac{1}{2x}$$
and $(2)$ to
$$\frac{\ \dfrac{2}{3x}\ }{x}=\dfrac{2}{3x}\cdot\dfrac{1}{x}=\dfrac{2\cdot 1}{3x\cdot x}=\dfrac{2}{3x^2}.$$
So by $(3)$ we have
$$\dfrac{\ \dfrac{1}{x}\ }{2} + \dfrac{\ \dfrac{2}{3x}\ }{x}=\dfrac{1}{2x}+\dfrac{2}{3x^2}=\dfrac{1\cdot 3x^2+2\times 2x}{2x\cdot 3x^2 }=\dfrac{3x^2+4x}{6x^3}=\dfrac{x(3x+4)}{x(6x^2)}=\dfrac{3x+4}{6x^2}.$$
For
$$\dfrac{x}{\; \dfrac{2}{x}\ } + \dfrac{\; \dfrac{1}{x}\; }{x}$$
we have
$$\dfrac{\; x\; }{\dfrac{2}{x}} + \dfrac{\; \dfrac{1}{x}\; }{x}=\dfrac{x\cdot x}{2} + \dfrac{1}{x\cdot x}=\dfrac{x^2}{2}+\cfrac{1}{x^2}=\dfrac{x^2\cdot x^2+2\cdot 1}{2\cdot x^2}=\dfrac{x^4+2}{2x^2}.$$
We can apply the property Division by a fraction is the same as multiplication by its inverse  or reciprocal to the following fraction 
$$\dfrac{\;\dfrac{a}{b}\;}{\dfrac{c}{d}}=\dfrac{a}{b}\cdot \dfrac{d}{c}=\dfrac{a\cdot d}{b\cdot c}=\dfrac{ad}{bc}\qquad (4).$$
A: Here is a start for the first one:
$$\frac{\frac{1}{x}}{2} + \frac{\frac{2}{3x}}{x} = \frac{x\frac{1}{x}}{2x} + \frac{2\frac{2}{3x}}{2x} 
= \frac{1}{2x} + \frac{\frac{4}{3x}}{2x} = \frac{1}{2x} + \frac{4}{3x}\frac{1}{2x} = \frac{1}{2x} + \frac{4}{6x^2}  = \frac{1}{2x} + \frac{2}{3x^2}$$
Now try to derive $\displaystyle\frac{3x + 4}{6x^2}$ from this.
A: Yet another strategy:
\begin{align}
\frac{\frac1x}2+\frac{\frac2{3x}}x&=\frac1{2x}+\frac2{3x^2}\\
&=\frac{3x}{6x^2}+\frac4{6x^2}=\frac{3x+4}{6x^2}\,.
\end{align}
What did I do? Given is the sum of two fractions, and I multiplied top-and-bottom of the first by $x$, and top-and-bottom of the second by $3x$. Second step, find the minimal common denominator, which is $6x^2$, and on each of your current fractions, multiply top-and-bottom by a suitable quantity to get the denominators equal. Now add.
