Solve 2a-1/a^2-2a+1+a^2-3a+2/4a^2-1

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

Evaluate

$$\frac{a^{4}+3a^{3}+5a^{2}-9a+3}{\left(a-1\right)^{2}\left(4a^{2}-1\right)}$$

Short Solution Steps

Factor a^{2}-2a+1. Factor 4a^{2}-1.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(a-1\right)^{2} and \left(2a-1\right)\left(2a+1\right) is \left(2a-1\right)\left(2a+1\right)\left(a-1\right)^{2}. Multiply \frac{2a-1}{\left(a-1\right)^{2}} times \frac{\left(2a-1\right)\left(2a+1\right)}{\left(2a-1\right)\left(2a+1\right)}. Multiply \frac{a^{2}-3a+2}{\left(2a-1\right)\left(2a+1\right)} times \frac{\left(a-1\right)^{2}}{\left(a-1\right)^{2}}.

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

Since \frac{\left(2a-1\right)\left(2a-1\right)\left(2a+1\right)}{\left(2a-1\right)\left(2a+1\right)\left(a-1\right)^{2}} and \frac{\left(a^{2}-3a+2\right)\left(a-1\right)^{2}}{\left(2a-1\right)\left(2a+1\right)\left(a-1\right)^{2}} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \left(2a-1\right)\left(2a-1\right)\left(2a+1\right)+\left(a^{2}-3a+2\right)\left(a-1\right)^{2}.

$$\frac{8a^{3}-2a-4a^{2}+1+a^{4}-2a^{3}+a^{2}-3a^{3}+6a^{2}-3a+2a^{2}-4a+2}{\left(2a-1\right)\left(2a+1\right)\left(a-1\right)^{2}}$$

Combine like terms in 8a^{3}-2a-4a^{2}+1+a^{4}-2a^{3}+a^{2}-3a^{3}+6a^{2}-3a+2a^{2}-4a+2.

$$\frac{3a^{3}-9a+5a^{2}+3+a^{4}}{\left(2a-1\right)\left(2a+1\right)\left(a-1\right)^{2}}$$

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

$$\frac{3a^{3}-9a+5a^{2}+3+a^{4}}{4a^{4}-8a^{3}+3a^{2}+2a-1}$$

Expand

$$\frac{a^{4}+3a^{3}+5a^{2}-9a+3}{\left(4a^{2}-1\right)\left(a^{2}-2a+1\right)}$$

Short Solution Steps

Factor a^{2}-2a+1. Factor 4a^{2}-1.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(a-1\right)^{2} and \left(2a-1\right)\left(2a+1\right) is \left(2a-1\right)\left(2a+1\right)\left(a-1\right)^{2}. Multiply \frac{2a-1}{\left(a-1\right)^{2}} times \frac{\left(2a-1\right)\left(2a+1\right)}{\left(2a-1\right)\left(2a+1\right)}. Multiply \frac{a^{2}-3a+2}{\left(2a-1\right)\left(2a+1\right)} times \frac{\left(a-1\right)^{2}}{\left(a-1\right)^{2}}.

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

Since \frac{\left(2a-1\right)\left(2a-1\right)\left(2a+1\right)}{\left(2a-1\right)\left(2a+1\right)\left(a-1\right)^{2}} and \frac{\left(a^{2}-3a+2\right)\left(a-1\right)^{2}}{\left(2a-1\right)\left(2a+1\right)\left(a-1\right)^{2}} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \left(2a-1\right)\left(2a-1\right)\left(2a+1\right)+\left(a^{2}-3a+2\right)\left(a-1\right)^{2}.

$$\frac{8a^{3}-2a-4a^{2}+1+a^{4}-2a^{3}+a^{2}-3a^{3}+6a^{2}-3a+2a^{2}-4a+2}{\left(2a-1\right)\left(2a+1\right)\left(a-1\right)^{2}}$$

Combine like terms in 8a^{3}-2a-4a^{2}+1+a^{4}-2a^{3}+a^{2}-3a^{3}+6a^{2}-3a+2a^{2}-4a+2.

$$\frac{3a^{3}-9a+5a^{2}+3+a^{4}}{\left(2a-1\right)\left(2a+1\right)\left(a-1\right)^{2}}$$

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

$$\frac{3a^{3}-9a+5a^{2}+3+a^{4}}{4a^{4}-8a^{3}+3a^{2}+2a-1}$$