Find $ \left(2x+1\right)^2 $ using formula.

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

where $ A = \color{blue}{ 2x } $ and $ B = \color{red}{ 1 }$.

$$ \begin{aligned}\left(2x+1\right)^2 = \color{blue}{\left( 2x \right)^2} +2 \cdot 2x \cdot 1 + \color{red}{1^2} = 4x^2+4x+1\end{aligned} $$

Simplify $ \dfrac{2x^2+x}{4x^2+4x+1} $ to $ \dfrac{x}{2x+1} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2x+1}$.

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

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

To subtract expressions with the same denominators, we subtract the numerators and write the result over the common denominator.

$$ \begin{aligned} \frac{x}{2x+1} - \frac{3}{2x+1} & = \frac{x}{\color{blue}{2x+1}} - \frac{3}{\color{blue}{2x+1}} =\frac{ x - 3 }{ \color{blue}{ 2x+1 }} \end{aligned} $$