Add $ \dfrac{1}{x} $ and $ \dfrac{2}{x-2} $ to get $ \dfrac{ \color{purple}{ 3x-2 } }{ x^2-2x }$.

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}{ x-2 }$ and the second by $\color{blue}{ x }$.

$$ \begin{aligned} \frac{1}{x} + \frac{2}{x-2} & = \frac{ 1 \cdot \color{blue}{ \left( x-2 \right) }}{ x \cdot \color{blue}{ \left( x-2 \right) }} + \frac{ 2 \cdot \color{blue}{ x }}{ \left( x-2 \right) \cdot \color{blue}{ x }} = \\[1ex] &=\frac{ \color{purple}{ x-2 } }{ x^2-2x } + \frac{ \color{purple}{ 2x } }{ x^2-2x }=\frac{ \color{purple}{ 3x-2 } }{ x^2-2x } \end{aligned} $$

Subtract $ \dfrac{2}{x+2} $ from $ \dfrac{3x-2}{x^2-2x} $ to get $ \dfrac{ \color{purple}{ x^2+8x-4 } }{ x^3-4x }$.

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

$$ \begin{aligned} \frac{3x-2}{x^2-2x} - \frac{2}{x+2} & = \frac{ \left( 3x-2 \right) \cdot \color{blue}{ \left( x+2 \right) }}{ \left( x^2-2x \right) \cdot \color{blue}{ \left( x+2 \right) }} - \frac{ 2 \cdot \color{blue}{ \left( x^2-2x \right) }}{ \left( x+2 \right) \cdot \color{blue}{ \left( x^2-2x \right) }} = \\[1ex] &=\frac{ \color{purple}{ 3x^2+6x-2x-4 } }{ x^3+ \cancel{2x^2} -\cancel{2x^2}-4x } - \frac{ \color{purple}{ 2x^2-4x } }{ x^3+ \cancel{2x^2} -\cancel{2x^2}-4x } = \\[1ex] &=\frac{ \color{purple}{ 3x^2+6x-2x-4 - \left( 2x^2-4x \right) } }{ x^3-4x }=\frac{ \color{purple}{ x^2+8x-4 } }{ x^3-4x } \end{aligned} $$