Question
Answer
$$\dfrac{1-9x^2}{9x^2+6x+1}=\dfrac{-3x+1}{3x+1}$$
Explanation
| $$ \begin{aligned}\frac{1-9x^2}{9x^2+6x+1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-3x+1}{3x+1}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{1-9x^2}{9x^2+6x+1} $ to $ \dfrac{-3x+1}{3x+1} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{3x+1}$. $$ \begin{aligned} \frac{1-9x^2}{9x^2+6x+1} & =\frac{ \left( -3x+1 \right) \cdot \color{blue}{ \left( 3x+1 \right) }}{ \left( 3x+1 \right) \cdot \color{blue}{ \left( 3x+1 \right) }} = \\[1ex] &= \frac{-3x+1}{3x+1} \end{aligned} $$ |
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