Add $ \dfrac{1}{x} $ and $ \dfrac{x}{2x+4} $ to get $ \dfrac{ \color{purple}{ x^2+2x+4 } }{ 2x^2+4x }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ 2x+4 }$ and the second by $\color{blue}{ x }$.

$$ \begin{aligned} \frac{1}{x} + \frac{x}{2x+4} & = \frac{ 1 \cdot \color{blue}{ \left( 2x+4 \right) }}{ x \cdot \color{blue}{ \left( 2x+4 \right) }} + \frac{ x \cdot \color{blue}{ x }}{ \left( 2x+4 \right) \cdot \color{blue}{ x }} = \\[1ex] &=\frac{ \color{purple}{ 2x+4 } }{ 2x^2+4x } + \frac{ \color{purple}{ x^2 } }{ 2x^2+4x }=\frac{ \color{purple}{ x^2+2x+4 } }{ 2x^2+4x } \end{aligned} $$

Subtract $ \dfrac{2}{x^2+2x} $ from $ \dfrac{x^2+2x+4}{2x^2+4x} $ to get $ \dfrac{ \color{purple}{ x^2+2x } }{ 2x^2+4x }$.

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{2}$.

$$ \begin{aligned} \frac{x^2+2x+4}{2x^2+4x} - \frac{2}{x^2+2x} & = \frac{ x^2+2x+4 }{ 2x^2+4x } - \frac{ 2 \cdot \color{blue}{ 2 }}{ \left( x^2+2x \right) \cdot \color{blue}{ 2 }} = \\[1ex] &=\frac{ \color{purple}{ x^2+2x+4 } }{ 2x^2+4x } - \frac{ \color{purple}{ 4 } }{ 2x^2+4x }=\frac{ \color{purple}{ x^2+2x } }{ 2x^2+4x } \end{aligned} $$

Simplify $ \dfrac{x^2+2x}{2x^2+4x} $ to $ \dfrac{1}{2} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x^2+2x}$.

$$ \begin{aligned} \frac{x^2+2x}{2x^2+4x} & =\frac{ 1 \cdot \color{blue}{ \left( x^2+2x \right) }}{ 2 \cdot \color{blue}{ \left( x^2+2x \right) }} = \\[1ex] &= \frac{1}{2} \end{aligned} $$