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

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.

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

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

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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