Multiply $ \dfrac{x-2}{2} $ by $ 4x-10 $ to get $ \dfrac{4x^2-18x+20}{2} $.

Step 1: Write $ 4x-10 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{x-2}{2} \cdot 4x-10 & \xlongequal{\text{Step 1}} \frac{x-2}{2} \cdot \frac{4x-10}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( x-2 \right) \cdot \left( 4x-10 \right) }{ 2 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 4x^2-10x-8x+20 }{ 2 } = \frac{4x^2-18x+20}{2} \end{aligned} $$

Subtract $ \dfrac{5}{6} $ from $ x $ to get $ \dfrac{ \color{purple}{ 6x-5 } }{ 6 }$.

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 first fraction by $\color{blue}{ 6 }$.

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

Multiply $25$ by $ \dfrac{x}{3} $ to get $ \dfrac{ 25x }{ 3 } $.

Step 1: Write $ 25 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 25 \cdot \frac{x}{3} & \xlongequal{\text{Step 1}} \frac{25}{\color{red}{1}} \cdot \frac{x}{3} \xlongequal{\text{Step 2}} \frac{ 25 \cdot x }{ 1 \cdot 3 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 25x }{ 3 } \end{aligned} $$

Multiply $ \dfrac{4x^2-18x+20}{2} $ by $ \dfrac{x-2}{3} $ to get $ \dfrac{4x^3-26x^2+56x-40}{6} $.

Step 1: Multiply numerators and denominators.

Step 2: Simplify numerator and denominator.

$$ \begin{aligned} \frac{4x^2-18x+20}{2} \cdot \frac{x-2}{3} & \xlongequal{\text{Step 1}} \frac{ \left( 4x^2-18x+20 \right) \cdot \left( x-2 \right) }{ 2 \cdot 3 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 4x^3-8x^2-18x^2+36x+20x-40 }{ 6 } = \frac{4x^3-26x^2+56x-40}{6} \end{aligned} $$

Multiply $ \dfrac{25}{2} $ by $ \dfrac{6x-5}{6} $ to get $ \dfrac{ 150x-125 }{ 12 } $.

Step 1: Multiply numerators and denominators.

Step 2: Simplify numerator and denominator.

$$ \begin{aligned} \frac{25}{2} \cdot \frac{6x-5}{6} & \xlongequal{\text{Step 1}} \frac{ 25 \cdot \left( 6x-5 \right) }{ 2 \cdot 6 } \xlongequal{\text{Step 2}} \frac{ 150x-125 }{ 12 } \end{aligned} $$

Multiply $25$ by $ \dfrac{x}{3} $ to get $ \dfrac{ 25x }{ 3 } $.

Step 1: Write $ 25 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 25 \cdot \frac{x}{3} & \xlongequal{\text{Step 1}} \frac{25}{\color{red}{1}} \cdot \frac{x}{3} \xlongequal{\text{Step 2}} \frac{ 25 \cdot x }{ 1 \cdot 3 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 25x }{ 3 } \end{aligned} $$

Subtract $ \dfrac{150x-125}{12} $ from $ \dfrac{4x^3-26x^2+56x-40}{6} $ to get $ \dfrac{ \color{purple}{ 8x^3-52x^2-38x+45 } }{ 12 }$.

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

$$ \begin{aligned} \frac{4x^3-26x^2+56x-40}{6} - \frac{150x-125}{12} & = \frac{ \left( 4x^3-26x^2+56x-40 \right) \cdot \color{blue}{ 2 }}{ 6 \cdot \color{blue}{ 2 }} - \frac{ 150x-125 }{ 12 } = \\[1ex] &=\frac{ \color{purple}{ 8x^3-52x^2+112x-80 } }{ 12 } - \frac{ \color{purple}{ 150x-125 } }{ 12 }=\frac{ \color{purple}{ 8x^3-52x^2+112x-80 - \left( 150x-125 \right) } }{ 12 } = \\[1ex] &=\frac{ \color{purple}{ 8x^3-52x^2-38x+45 } }{ 12 } \end{aligned} $$

Multiply $25$ by $ \dfrac{x}{3} $ to get $ \dfrac{ 25x }{ 3 } $.

Step 1: Write $ 25 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 25 \cdot \frac{x}{3} & \xlongequal{\text{Step 1}} \frac{25}{\color{red}{1}} \cdot \frac{x}{3} \xlongequal{\text{Step 2}} \frac{ 25 \cdot x }{ 1 \cdot 3 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 25x }{ 3 } \end{aligned} $$

Add $ \dfrac{8x^3-52x^2-38x+45}{12} $ and $ \dfrac{25x}{3} $ to get $ \dfrac{ \color{purple}{ 8x^3-52x^2+62x+45 } }{ 12 }$.

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

$$ \begin{aligned} \frac{8x^3-52x^2-38x+45}{12} + \frac{25x}{3} & = \frac{ 8x^3-52x^2-38x+45 }{ 12 } + \frac{ 25x \cdot \color{blue}{ 4 }}{ 3 \cdot \color{blue}{ 4 }} = \\[1ex] &=\frac{ \color{purple}{ 8x^3-52x^2-38x+45 } }{ 12 } + \frac{ \color{purple}{ 100x } }{ 12 }=\frac{ \color{purple}{ 8x^3-52x^2+62x+45 } }{ 12 } \end{aligned} $$