Add $x^2-3x$ and $ \dfrac{2}{x^2} $ to get $ \dfrac{ \color{purple}{ x^4-3x^3+2 } }{ x^2 }$.

Step 1: Write $ x^2-3x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

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

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

Subtract $1$ from $ \dfrac{x^4-3x^3+2}{x^2} $ to get $ \dfrac{ \color{purple}{ x^4-3x^3-x^2+2 } }{ x^2 }$.

Step 1: Write $ 1 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

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

$$ \begin{aligned} \frac{x^4-3x^3+2}{x^2} -1 & \xlongequal{\text{Step 1}} \frac{x^4-3x^3+2}{x^2} - \frac{1}{\color{red}{1}} = \frac{ x^4-3x^3+2 }{ x^2 } - \frac{ 1 \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x^4-3x^3+2 } }{ x^2 } - \frac{ \color{purple}{ x^2 } }{ x^2 }=\frac{ \color{purple}{ x^4-3x^3-x^2+2 } }{ x^2 } \end{aligned} $$