Add $1$ and $ \dfrac{x^2-2x+1}{1} $ to get $ \dfrac{x^2-2x+2}{1} $.

Step 1: Write $ 1 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To add expressions with the same denominators, we add the numerators and write the result over the common denominator.

$$ \begin{aligned} 1+ \frac{x^2-2x+1}{1} & \xlongequal{\text{Step 1}} \frac{1}{\color{red}{1}} + \frac{x^2-2x+1}{1} \xlongequal{\text{Step 2}} \frac{1}{\color{blue}{1}} + \frac{x^2-2x+1}{\color{blue}{1}} = \\[1ex] &=\frac{ 1 + \left( x^2-2x+1 \right) }{ \color{blue}{ 1 }}= \frac{x^2-2x+2}{1} \end{aligned} $$

Divide $1$ by $ \dfrac{x^2-2x+2}{1} $ to get $ \dfrac{ 1 }{ x^2-2x+2 } $.

Step 1: To divide rational expressions, multiply the first fraction by the reciprocal of the second fraction.

Step 2: Write $ 1 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

$$ \begin{aligned} \frac{1}{ \frac{\color{blue}{x^2-2x+2}}{\color{blue}{1}} } & \xlongequal{\text{Step 1}} 1 \cdot \frac{\color{blue}{1}}{\color{blue}{x^2-2x+2}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{1}{\color{red}{1}} \cdot \frac{1}{x^2-2x+2} \xlongequal{\text{Step 3}} \frac{ 1 \cdot 1 }{ 1 \cdot \left( x^2-2x+2 \right) } = \\[1ex] & \xlongequal{\text{Step 4}} \frac{ 1 }{ x^2-2x+2 } \end{aligned} $$