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

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

Step 2: To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ x^2 }$.

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

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

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.

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