Find $ \left(x+2\right)^2 $ using formula.

$$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$

where $ A = \color{blue}{ x } $ and $ B = \color{red}{ 2 }$.

$$ \begin{aligned}\left(x+2\right)^2 = \color{blue}{x^2} +2 \cdot x \cdot 2 + \color{red}{2^2} = x^2+4x+4\end{aligned} $$

Multiply $3$ by $ \dfrac{x}{x^2+4x+4} $ to get $ \dfrac{ 3x }{ x^2+4x+4 } $.

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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

Add $ \dfrac{7}{x+2} $ and $ \dfrac{3x}{x^2+4x+4} $ to get $ \dfrac{ \color{purple}{ 10x+14 } }{ x^2+4x+4 }$.

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

$$ \begin{aligned} \frac{7}{x+2} + \frac{3x}{x^2+4x+4} & = \frac{ 7 \cdot \color{blue}{ \left( x+2 \right) }}{ \left( x+2 \right) \cdot \color{blue}{ \left( x+2 \right) }} + \frac{ 3x }{ x^2+4x+4 } = \\[1ex] &=\frac{ \color{purple}{ 7x+14 } }{ x^2+2x+2x+4 } + \frac{ \color{purple}{ 3x } }{ x^2+2x+2x+4 }=\frac{ \color{purple}{ 10x+14 } }{ x^2+4x+4 } \end{aligned} $$

Subtract $ \dfrac{5}{x} $ from $ \dfrac{10x+14}{x^2+4x+4} $ to get $ \dfrac{ \color{purple}{ 5x^2-6x-20 } }{ x^3+4x^2+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 }$ and the second by $\color{blue}{ x^2+4x+4 }$.

$$ \begin{aligned} \frac{10x+14}{x^2+4x+4} - \frac{5}{x} & = \frac{ \left( 10x+14 \right) \cdot \color{blue}{ x }}{ \left( x^2+4x+4 \right) \cdot \color{blue}{ x }} - \frac{ 5 \cdot \color{blue}{ \left( x^2+4x+4 \right) }}{ x \cdot \color{blue}{ \left( x^2+4x+4 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 10x^2+14x } }{ x^3+4x^2+4x } - \frac{ \color{purple}{ 5x^2+20x+20 } }{ x^3+4x^2+4x } = \\[1ex] &=\frac{ \color{purple}{ 10x^2+14x - \left( 5x^2+20x+20 \right) } }{ x^3+4x^2+4x }=\frac{ \color{purple}{ 5x^2-6x-20 } }{ x^3+4x^2+4x } \end{aligned} $$