Answer
$$\frac{(x-1)^2}{2-2x}=\frac{x-1}{-2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(x-1)^2}{2-2x}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x^2-2x+1}{2-2x} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x-1}{-2}\end{aligned} $$ | |
| ① | Find $ \left(x-1\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ x } $ and $ B = \color{red}{ 1 }$. $$ \begin{aligned}\left(x-1\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 1 + \color{red}{1^2} = x^2-2x+1\end{aligned} $$ |
| ② | Simplify $ \dfrac{x^2-2x+1}{2-2x} $ to $ \dfrac{x-1}{-2} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-1}$. $$ \begin{aligned} \frac{x^2-2x+1}{2-2x} & =\frac{ \left( x-1 \right) \cdot \color{blue}{ \left( x-1 \right) }}{ \left( -2 \right) \cdot \color{blue}{ \left( x-1 \right) }} = \\[1ex] &= \frac{x-1}{-2} \end{aligned} $$ |
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