Question
Answer
$$\dfrac{500b+200ab+20a^2b}{125b-5a^2b}=\dfrac{4a+20}{-a+5}$$
Explanation
| $$ \begin{aligned}\frac{500b+200ab+20a^2b}{125b-5a^2b}& \xlongequal{ }\frac{100+40a+4a^2}{25-a^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4a+20}{-a+5}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{100+40a+4a^2}{25-a^2} $ to $ \dfrac{4a+20}{-a+5} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{a+5}$. $$ \begin{aligned} \frac{100+40a+4a^2}{25-a^2} & =\frac{ \left( 4a+20 \right) \cdot \color{blue}{ \left( a+5 \right) }}{ \left( -a+5 \right) \cdot \color{blue}{ \left( a+5 \right) }} = \\[1ex] &= \frac{4a+20}{-a+5} \end{aligned} $$ |
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