Solve (2m+1/2m-1-2m-1/2m+1):4/10

$$( \frac { 2 m + 1 } { 2 m - 1 } - \frac { 2 m - 1 } { 2 m + 1 } ) : \frac { 4 } { 10 }$$

Evaluate

$$\frac{20m}{4m^{2}-1}$$

Short Solution Steps

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

$$\frac{\frac{\left(2m+1\right)\left(2m+1\right)}{\left(2m-1\right)\left(2m+1\right)}-\frac{\left(2m-1\right)\left(2m-1\right)}{\left(2m-1\right)\left(2m+1\right)}}{\frac{4}{10}}$$

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

$$\frac{\frac{\left(2m+1\right)\left(2m+1\right)-\left(2m-1\right)\left(2m-1\right)}{\left(2m-1\right)\left(2m+1\right)}}{\frac{4}{10}}$$

Do the multiplications in \left(2m+1\right)\left(2m+1\right)-\left(2m-1\right)\left(2m-1\right).

$$\frac{\frac{4m^{2}+2m+2m+1-4m^{2}+2m+2m-1}{\left(2m-1\right)\left(2m+1\right)}}{\frac{4}{10}}$$

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

$$\frac{\frac{8m}{\left(2m-1\right)\left(2m+1\right)}}{\frac{4}{10}}$$

Reduce the fraction \frac{4}{10} to lowest terms by extracting and canceling out 2.

$$\frac{\frac{8m}{\left(2m-1\right)\left(2m+1\right)}}{\frac{2}{5}}$$

Divide \frac{8m}{\left(2m-1\right)\left(2m+1\right)} by \frac{2}{5} by multiplying \frac{8m}{\left(2m-1\right)\left(2m+1\right)} by the reciprocal of \frac{2}{5}.

$$\frac{8m\times 5}{\left(2m-1\right)\left(2m+1\right)\times 2}$$

Cancel out 2 in both numerator and denominator.

$$\frac{4\times 5m}{\left(2m-1\right)\left(2m+1\right)}$$

Multiply 4 and 5 to get 20.

$$\frac{20m}{\left(2m-1\right)\left(2m+1\right)}$$

Consider \left(2m-1\right)\left(2m+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{20m}{\left(2m\right)^{2}-1^{2}}$$

Expand \left(2m\right)^{2}.

$$\frac{20m}{2^{2}m^{2}-1^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{20m}{4m^{2}-1^{2}}$$

Calculate 1 to the power of 2 and get 1.

$$\frac{20m}{4m^{2}-1}$$

Expand

$$\frac{20m}{4m^{2}-1}$$

Short Solution Steps

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

$$\frac{\frac{\left(2m+1\right)\left(2m+1\right)}{\left(2m-1\right)\left(2m+1\right)}-\frac{\left(2m-1\right)\left(2m-1\right)}{\left(2m-1\right)\left(2m+1\right)}}{\frac{4}{10}}$$

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

$$\frac{\frac{\left(2m+1\right)\left(2m+1\right)-\left(2m-1\right)\left(2m-1\right)}{\left(2m-1\right)\left(2m+1\right)}}{\frac{4}{10}}$$

Do the multiplications in \left(2m+1\right)\left(2m+1\right)-\left(2m-1\right)\left(2m-1\right).

$$\frac{\frac{4m^{2}+2m+2m+1-4m^{2}+2m+2m-1}{\left(2m-1\right)\left(2m+1\right)}}{\frac{4}{10}}$$

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

$$\frac{\frac{8m}{\left(2m-1\right)\left(2m+1\right)}}{\frac{4}{10}}$$

Reduce the fraction \frac{4}{10} to lowest terms by extracting and canceling out 2.

$$\frac{\frac{8m}{\left(2m-1\right)\left(2m+1\right)}}{\frac{2}{5}}$$

Divide \frac{8m}{\left(2m-1\right)\left(2m+1\right)} by \frac{2}{5} by multiplying \frac{8m}{\left(2m-1\right)\left(2m+1\right)} by the reciprocal of \frac{2}{5}.

$$\frac{8m\times 5}{\left(2m-1\right)\left(2m+1\right)\times 2}$$

Cancel out 2 in both numerator and denominator.

$$\frac{4\times 5m}{\left(2m-1\right)\left(2m+1\right)}$$

Multiply 4 and 5 to get 20.

$$\frac{20m}{\left(2m-1\right)\left(2m+1\right)}$$

Consider \left(2m-1\right)\left(2m+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{20m}{\left(2m\right)^{2}-1^{2}}$$

Expand \left(2m\right)^{2}.

$$\frac{20m}{2^{2}m^{2}-1^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{20m}{4m^{2}-1^{2}}$$

Calculate 1 to the power of 2 and get 1.

$$\frac{20m}{4m^{2}-1}$$