Subtract $1$ from $ \dfrac{8}{x} $ to get $ \dfrac{ \color{purple}{ -x+8 } }{ x }$.

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 $\color{blue}{x}$.

$$ \begin{aligned} \frac{8}{x} -1 & \xlongequal{\text{Step 1}} \frac{8}{x} - \frac{1}{\color{red}{1}} = \frac{ 8 }{ x } - \frac{ 1 \cdot \color{blue}{ x }}{ 1 \cdot \color{blue}{ x }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 8 } }{ x } - \frac{ \color{purple}{ x } }{ x }=\frac{ \color{purple}{ -x+8 } }{ x } \end{aligned} $$

Add $ \dfrac{-x+8}{x} $ and $ 5x $ to get $ \dfrac{ \color{purple}{ 5x^2-x+8 } }{ x }$.

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

$$ \begin{aligned} \frac{-x+8}{x} +5x & \xlongequal{\text{Step 1}} \frac{-x+8}{x} + \frac{5x}{\color{red}{1}} = \frac{ -x+8 }{ x } + \frac{ 5x \cdot \color{blue}{ x }}{ 1 \cdot \color{blue}{ x }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ -x+8 } }{ x } + \frac{ \color{purple}{ 5x^2 } }{ x }=\frac{ \color{purple}{ 5x^2-x+8 } }{ x } \end{aligned} $$

Add $ \dfrac{5x^2-x+8}{x} $ and $ \dfrac{8}{x^2} $ to get $ \dfrac{ \color{purple}{ 5x^4-x^3+8x^2+8x } }{ 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^2 }$ and the second by $\color{blue}{ x }$.

$$ \begin{aligned} \frac{5x^2-x+8}{x} + \frac{8}{x^2} & = \frac{ \left( 5x^2-x+8 \right) \cdot \color{blue}{ x^2 }}{ x \cdot \color{blue}{ x^2 }} + \frac{ 8 \cdot \color{blue}{ x }}{ x^2 \cdot \color{blue}{ x }} = \\[1ex] &=\frac{ \color{purple}{ 5x^4-x^3+8x^2 } }{ x^3 } + \frac{ \color{purple}{ 8x } }{ x^3 }=\frac{ \color{purple}{ 5x^4-x^3+8x^2+8x } }{ x^3 } \end{aligned} $$

Add $ \dfrac{5x^4-x^3+8x^2+8x}{x^3} $ and $ 4x $ to get $ \dfrac{ \color{purple}{ 9x^4-x^3+8x^2+8x } }{ x^3 }$.

Step 1: Write $ 4x $ 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 $\color{blue}{x^3}$.

$$ \begin{aligned} \frac{5x^4-x^3+8x^2+8x}{x^3} +4x & \xlongequal{\text{Step 1}} \frac{5x^4-x^3+8x^2+8x}{x^3} + \frac{4x}{\color{red}{1}} = \frac{ 5x^4-x^3+8x^2+8x }{ x^3 } + \frac{ 4x \cdot \color{blue}{ x^3 }}{ 1 \cdot \color{blue}{ x^3 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 5x^4-x^3+8x^2+8x } }{ x^3 } + \frac{ \color{purple}{ 4x^4 } }{ x^3 }=\frac{ \color{purple}{ 9x^4-x^3+8x^2+8x } }{ x^3 } \end{aligned} $$

Subtract $5$ from $ \dfrac{9x^4-x^3+8x^2+8x}{x^3} $ to get $ \dfrac{ \color{purple}{ 9x^4-6x^3+8x^2+8x } }{ x^3 }$.

Step 1: Write $ 5 $ 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 $\color{blue}{x^3}$.

$$ \begin{aligned} \frac{9x^4-x^3+8x^2+8x}{x^3} -5 & \xlongequal{\text{Step 1}} \frac{9x^4-x^3+8x^2+8x}{x^3} - \frac{5}{\color{red}{1}} = \frac{ 9x^4-x^3+8x^2+8x }{ x^3 } - \frac{ 5 \cdot \color{blue}{ x^3 }}{ 1 \cdot \color{blue}{ x^3 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 9x^4-x^3+8x^2+8x } }{ x^3 } - \frac{ \color{purple}{ 5x^3 } }{ x^3 }=\frac{ \color{purple}{ 9x^4-6x^3+8x^2+8x } }{ x^3 } \end{aligned} $$