Answer
$$\frac{x+9}{x^2-81}=\frac{1}{x-9}$$
Explanation
| $$ \begin{aligned}\frac{x+9}{x^2-81}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{x-9}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{x+9}{x^2-81} $ to $ \dfrac{1}{x-9} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+9}$. $$ \begin{aligned} \frac{x+9}{x^2-81} & =\frac{ 1 \cdot \color{blue}{ \left( x+9 \right) }}{ \left( x-9 \right) \cdot \color{blue}{ \left( x+9 \right) }} = \\[1ex] &= \frac{1}{x-9} \end{aligned} $$ |
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Simplify Rational Expressions