Add $3x$ and $ \dfrac{9}{x^2} $ to get $ \dfrac{ \color{purple}{ 3x^3+9 } }{ x^2 }$.

Step 1: Write $ 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 $\color{blue}{ x^2 }$.

$$ \begin{aligned} 3x+ \frac{9}{x^2} & \xlongequal{\text{Step 1}} \frac{3x}{\color{red}{1}} + \frac{9}{x^2} = \frac{ 3x \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} + \frac{ 9 }{ x^2 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 3x^3 } }{ x^2 } + \frac{ \color{purple}{ 9 } }{ x^2 }=\frac{ \color{purple}{ 3x^3+9 } }{ x^2 } \end{aligned} $$

Subtract $x$ from $ \dfrac{3x^3+9}{x^2} $ to get $ \dfrac{ \color{purple}{ 2x^3+9 } }{ x^2 }$.

Step 1: Write $ x $ 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 $\color{blue}{x^2}$.

$$ \begin{aligned} \frac{3x^3+9}{x^2} -x & \xlongequal{\text{Step 1}} \frac{3x^3+9}{x^2} - \frac{x}{\color{red}{1}} = \frac{ 3x^3+9 }{ x^2 } - \frac{ x \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 3x^3+9 } }{ x^2 } - \frac{ \color{purple}{ x^3 } }{ x^2 }=\frac{ \color{purple}{ 2x^3+9 } }{ x^2 } \end{aligned} $$

Subtract $2x^2$ from $ \dfrac{2x^3+9}{x^2} $ to get $ \dfrac{ \color{purple}{ -2x^4+2x^3+9 } }{ x^2 }$.

Step 1: Write $ 2x^2 $ 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 $\color{blue}{x^2}$.

$$ \begin{aligned} \frac{2x^3+9}{x^2} -2x^2 & \xlongequal{\text{Step 1}} \frac{2x^3+9}{x^2} - \frac{2x^2}{\color{red}{1}} = \frac{ 2x^3+9 }{ x^2 } - \frac{ 2x^2 \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 2x^3+9 } }{ x^2 } - \frac{ \color{purple}{ 2x^4 } }{ x^2 }=\frac{ \color{purple}{ -2x^4+2x^3+9 } }{ x^2 } \end{aligned} $$

Subtract $5x$ from $ \dfrac{-2x^4+2x^3+9}{x^2} $ to get $ \dfrac{ \color{purple}{ -2x^4-3x^3+9 } }{ x^2 }$.

Step 1: Write $ 5x $ 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 $\color{blue}{x^2}$.

$$ \begin{aligned} \frac{-2x^4+2x^3+9}{x^2} -5x & \xlongequal{\text{Step 1}} \frac{-2x^4+2x^3+9}{x^2} - \frac{5x}{\color{red}{1}} = \frac{ -2x^4+2x^3+9 }{ x^2 } - \frac{ 5x \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ -2x^4+2x^3+9 } }{ x^2 } - \frac{ \color{purple}{ 5x^3 } }{ x^2 }=\frac{ \color{purple}{ -2x^4-3x^3+9 } }{ x^2 } \end{aligned} $$

Add $ \dfrac{-2x^4-3x^3+9}{x^2} $ and $ \dfrac{6}{x} $ to get $ \dfrac{ \color{purple}{ -2x^5-3x^4+6x^2+9x } }{ x^3 }$.

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

$$ \begin{aligned} \frac{-2x^4-3x^3+9}{x^2} + \frac{6}{x} & = \frac{ \left( -2x^4-3x^3+9 \right) \cdot \color{blue}{ x }}{ x^2 \cdot \color{blue}{ x }} + \frac{ 6 \cdot \color{blue}{ x^2 }}{ x \cdot \color{blue}{ x^2 }} = \\[1ex] &=\frac{ \color{purple}{ -2x^5-3x^4+9x } }{ x^3 } + \frac{ \color{purple}{ 6x^2 } }{ x^3 }=\frac{ \color{purple}{ -2x^5-3x^4+6x^2+9x } }{ x^3 } \end{aligned} $$

Add $ \dfrac{-2x^5-3x^4+6x^2+9x}{x^3} $ and $ 3 $ to get $ \dfrac{ \color{purple}{ -2x^5-3x^4+3x^3+6x^2+9x } }{ x^3 }$.

Step 1: Write $ 3 $ 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 second fraction by $\color{blue}{x^3}$.

$$ \begin{aligned} \frac{-2x^5-3x^4+6x^2+9x}{x^3} +3 & \xlongequal{\text{Step 1}} \frac{-2x^5-3x^4+6x^2+9x}{x^3} + \frac{3}{\color{red}{1}} = \frac{ -2x^5-3x^4+6x^2+9x }{ x^3 } + \frac{ 3 \cdot \color{blue}{ x^3 }}{ 1 \cdot \color{blue}{ x^3 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ -2x^5-3x^4+6x^2+9x } }{ x^3 } + \frac{ \color{purple}{ 3x^3 } }{ x^3 }=\frac{ \color{purple}{ -2x^5-3x^4+3x^3+6x^2+9x } }{ x^3 } \end{aligned} $$