| $$ \begin{aligned}\frac{4x^2-1}{2x-1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2x+1\end{aligned} $$ | |
| ① | Simplify $ \dfrac{4x^2-1}{2x-1} $ to $ 2x+1$. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2x-1}$. $$ \begin{aligned} \frac{4x^2-1}{2x-1} & =\frac{ \left( 2x+1 \right) \cdot \color{blue}{ \left( 2x-1 \right) }}{ 1 \cdot \color{blue}{ \left( 2x-1 \right) }} = \\[1ex] &= \frac{2x+1}{1} =2x+1 \end{aligned} $$ |