Answer
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{x^2+1}{2x^2-4x+2}+\frac{x}{(x-1)^2}-\frac{x+1}{x^2-2x+1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x^2+1}{2x^2-4x+2}+\frac{x}{x^2-2x+1}-\frac{x+1}{x^2-2x+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x^2+2x+1}{2x^2-4x+2}-\frac{x+1}{x^2-2x+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{x^2-1}{2x^2-4x+2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{x+1}{2x-2}\end{aligned} $$ | |
| 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 | Add $ \dfrac{x^2+1}{2x^2-4x+2} $ and $ \dfrac{x}{x^2-2x+1} $ to get $ \dfrac{ \color{purple}{ x^2+2x+1 } }{ 2x^2-4x+2 }$. To add raitonal expressions, both fractions must have the same denominator. |
| ③ | Subtract $ \dfrac{x+1}{x^2-2x+1} $ from $ \dfrac{x^2+2x+1}{2x^2-4x+2} $ to get $ \dfrac{ \color{purple}{ x^2-1 } }{ 2x^2-4x+2 }$. To subtract raitonal expressions, both fractions must have the same denominator. |
| ④ | Simplify $ \dfrac{x^2-1}{2x^2-4x+2} $ to $ \dfrac{x+1}{2x-2} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-1}$. $$ \begin{aligned} \frac{x^2-1}{2x^2-4x+2} & =\frac{ \left( x+1 \right) \cdot \color{blue}{ \left( x-1 \right) }}{ \left( 2x-2 \right) \cdot \color{blue}{ \left( x-1 \right) }} = \\[1ex] &= \frac{x+1}{2x-2} \end{aligned} $$ |