Divide $ \dfrac{x^2-5x+4}{x^2} $ by $ \dfrac{x-1}{x} $ to get $ \dfrac{ x^2-4x }{ x^2 } $.

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

Step 2: Factor numerators and denominators.

Step 3: Cancel common factors.

Step 4: Multiply numerators and denominators.

Step 5: Simplify numerator and denominator.

$$ \begin{aligned} \frac{ \frac{x^2-5x+4}{x^2} }{ \frac{\color{blue}{x-1}}{\color{blue}{x}} } & \xlongequal{\text{Step 1}} \frac{x^2-5x+4}{x^2} \cdot \frac{\color{blue}{x}}{\color{blue}{x-1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( x-4 \right) \cdot \color{blue}{ \left( x-1 \right) } }{ x^2 } \cdot \frac{ x }{ 1 \cdot \color{blue}{ \left( x-1 \right) } } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x-4 }{ x^2 } \cdot \frac{ x }{ 1 } \xlongequal{\text{Step 4}} \frac{ \left( x-4 \right) \cdot x }{ x^2 \cdot 1 } \xlongequal{\text{Step 5}} \frac{ x^2-4x }{ x^2 } \end{aligned} $$