Solve 1-2d-1/4d^2-2d+1-2d/2d+1

$$1 - \frac { 2 d - 1 } { 4 d ^ { 2 } - 2 d + 1 } - \frac { 2 d } { 2 d + 1 }$$

Evaluate

$$\frac{2\left(1-d\right)}{8d^{3}+1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{4d^{2}-2d+1}{4d^{2}-2d+1}.

$$\frac{4d^{2}-2d+1}{4d^{2}-2d+1}-\frac{2d-1}{4d^{2}-2d+1}-\frac{2d}{2d+1}$$

Since \frac{4d^{2}-2d+1}{4d^{2}-2d+1} and \frac{2d-1}{4d^{2}-2d+1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4d^{2}-2d+1-\left(2d-1\right)}{4d^{2}-2d+1}-\frac{2d}{2d+1}$$

Do the multiplications in 4d^{2}-2d+1-\left(2d-1\right).

$$\frac{4d^{2}-2d+1-2d+1}{4d^{2}-2d+1}-\frac{2d}{2d+1}$$

Combine like terms in 4d^{2}-2d+1-2d+1.

$$\frac{4d^{2}-4d+2}{4d^{2}-2d+1}-\frac{2d}{2d+1}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4d^{2}-2d+1 and 2d+1 is \left(2d+1\right)\left(4d^{2}-2d+1\right). Multiply \frac{4d^{2}-4d+2}{4d^{2}-2d+1} times \frac{2d+1}{2d+1}. Multiply \frac{2d}{2d+1} times \frac{4d^{2}-2d+1}{4d^{2}-2d+1}.

$$\frac{\left(4d^{2}-4d+2\right)\left(2d+1\right)}{\left(2d+1\right)\left(4d^{2}-2d+1\right)}-\frac{2d\left(4d^{2}-2d+1\right)}{\left(2d+1\right)\left(4d^{2}-2d+1\right)}$$

Since \frac{\left(4d^{2}-4d+2\right)\left(2d+1\right)}{\left(2d+1\right)\left(4d^{2}-2d+1\right)} and \frac{2d\left(4d^{2}-2d+1\right)}{\left(2d+1\right)\left(4d^{2}-2d+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(4d^{2}-4d+2\right)\left(2d+1\right)-2d\left(4d^{2}-2d+1\right)}{\left(2d+1\right)\left(4d^{2}-2d+1\right)}$$

Do the multiplications in \left(4d^{2}-4d+2\right)\left(2d+1\right)-2d\left(4d^{2}-2d+1\right).

$$\frac{8d^{3}+4d^{2}-8d^{2}-4d+4d+2-8d^{3}+4d^{2}-2d}{\left(2d+1\right)\left(4d^{2}-2d+1\right)}$$

Combine like terms in 8d^{3}+4d^{2}-8d^{2}-4d+4d+2-8d^{3}+4d^{2}-2d.

$$\frac{-2d+2}{\left(2d+1\right)\left(4d^{2}-2d+1\right)}$$

Expand \left(2d+1\right)\left(4d^{2}-2d+1\right).

$$\frac{-2d+2}{8d^{3}+1}$$

Expand

$$-\frac{2\left(d-1\right)}{8d^{3}+1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{4d^{2}-2d+1}{4d^{2}-2d+1}.

$$\frac{4d^{2}-2d+1}{4d^{2}-2d+1}-\frac{2d-1}{4d^{2}-2d+1}-\frac{2d}{2d+1}$$

Since \frac{4d^{2}-2d+1}{4d^{2}-2d+1} and \frac{2d-1}{4d^{2}-2d+1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4d^{2}-2d+1-\left(2d-1\right)}{4d^{2}-2d+1}-\frac{2d}{2d+1}$$

Do the multiplications in 4d^{2}-2d+1-\left(2d-1\right).

$$\frac{4d^{2}-2d+1-2d+1}{4d^{2}-2d+1}-\frac{2d}{2d+1}$$

Combine like terms in 4d^{2}-2d+1-2d+1.

$$\frac{4d^{2}-4d+2}{4d^{2}-2d+1}-\frac{2d}{2d+1}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4d^{2}-2d+1 and 2d+1 is \left(2d+1\right)\left(4d^{2}-2d+1\right). Multiply \frac{4d^{2}-4d+2}{4d^{2}-2d+1} times \frac{2d+1}{2d+1}. Multiply \frac{2d}{2d+1} times \frac{4d^{2}-2d+1}{4d^{2}-2d+1}.

$$\frac{\left(4d^{2}-4d+2\right)\left(2d+1\right)}{\left(2d+1\right)\left(4d^{2}-2d+1\right)}-\frac{2d\left(4d^{2}-2d+1\right)}{\left(2d+1\right)\left(4d^{2}-2d+1\right)}$$

Since \frac{\left(4d^{2}-4d+2\right)\left(2d+1\right)}{\left(2d+1\right)\left(4d^{2}-2d+1\right)} and \frac{2d\left(4d^{2}-2d+1\right)}{\left(2d+1\right)\left(4d^{2}-2d+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(4d^{2}-4d+2\right)\left(2d+1\right)-2d\left(4d^{2}-2d+1\right)}{\left(2d+1\right)\left(4d^{2}-2d+1\right)}$$

Do the multiplications in \left(4d^{2}-4d+2\right)\left(2d+1\right)-2d\left(4d^{2}-2d+1\right).

$$\frac{8d^{3}+4d^{2}-8d^{2}-4d+4d+2-8d^{3}+4d^{2}-2d}{\left(2d+1\right)\left(4d^{2}-2d+1\right)}$$

Combine like terms in 8d^{3}+4d^{2}-8d^{2}-4d+4d+2-8d^{3}+4d^{2}-2d.

$$\frac{-2d+2}{\left(2d+1\right)\left(4d^{2}-2d+1\right)}$$

Expand \left(2d+1\right)\left(4d^{2}-2d+1\right).

$$\frac{-2d+2}{8d^{3}+1}$$