Solve m/2m^2+m+1-1/2m^2+m-1

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

Evaluate

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

Short Solution Steps

Factor 2m^{2}+m-1.

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

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

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

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

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

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

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

Combine like terms in 2m^{3}+2m^{2}-m^{2}-m-2m^{2}-m-1.

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

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

$$\frac{2m^{3}-m^{2}-2m-1}{4m^{4}+4m^{3}+m^{2}-1}$$

Differentiate w.r.t. m

$$\frac{2\left(1+2m+4m^{2}+16m^{3}+15m^{4}+4m^{5}-4m^{6}\right)}{\left(\left(2m-1\right)\left(m+1\right)\left(2m^{2}+m+1\right)\right)^{2}}$$