Question
Answer
$$\dfrac{4a^2b-2ab^2}{(2a-b)^2}=\dfrac{4a^2b-2ab^2}{4a^2-4ab+b^2}$$
Explanation
| $$ \begin{aligned}\frac{4a^2b-2ab^2}{(2a-b)^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4a^2b-2ab^2}{4a^2-4ab+b^2}\end{aligned} $$ | |
| ① | Find $ \left(2a-b\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 2a } $ and $ B = \color{red}{ b }$. $$ \begin{aligned}\left(2a-b\right)^2 = \color{blue}{\left( 2a \right)^2} -2 \cdot 2a \cdot b + \color{red}{b^2} = 4a^2-4ab+b^2\end{aligned} $$ |
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