Solve f(x)=(3a^2+b^2/a^2-b^2-2a/a-b)div2/a+b

$$f ( x ) = ( \frac { 3 a ^ { 2 } + b ^ { 2 } } { a ^ { 2 } - b ^ { 2 } } - \frac { 2 a } { a - b } ) \div \frac { 2 } { a + b }$$

Evaluate

$$\frac{a-b}{2}$$

Short Solution Steps

Factor a^{2}-b^{2}.

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

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

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

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

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

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

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

Combine like terms in 3a^{2}+b^{2}-2a^{2}-2ab.

$$\frac{\frac{a^{2}-2ab+b^{2}}{\left(a+b\right)\left(a-b\right)}}{\frac{2}{a+b}}$$

Factor the expressions that are not already factored in \frac{a^{2}-2ab+b^{2}}{\left(a+b\right)\left(a-b\right)}.

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

Cancel out a-b in both numerator and denominator.

$$\frac{\frac{a-b}{a+b}}{\frac{2}{a+b}}$$

Divide \frac{a-b}{a+b} by \frac{2}{a+b} by multiplying \frac{a-b}{a+b} by the reciprocal of \frac{2}{a+b}.

$$\frac{\left(a-b\right)\left(a+b\right)}{\left(a+b\right)\times 2}$$

Cancel out a+b in both numerator and denominator.

$$\frac{a-b}{2}$$

Expand

$$\frac{a-b}{2}$$

Short Solution Steps

Factor a^{2}-b^{2}.

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

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

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

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

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

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

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

Combine like terms in 3a^{2}+b^{2}-2a^{2}-2ab.

$$\frac{\frac{a^{2}-2ab+b^{2}}{\left(a+b\right)\left(a-b\right)}}{\frac{2}{a+b}}$$

Factor the expressions that are not already factored in \frac{a^{2}-2ab+b^{2}}{\left(a+b\right)\left(a-b\right)}.

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

Cancel out a-b in both numerator and denominator.

$$\frac{\frac{a-b}{a+b}}{\frac{2}{a+b}}$$

Divide \frac{a-b}{a+b} by \frac{2}{a+b} by multiplying \frac{a-b}{a+b} by the reciprocal of \frac{2}{a+b}.

$$\frac{\left(a-b\right)\left(a+b\right)}{\left(a+b\right)\times 2}$$

Cancel out a+b in both numerator and denominator.

$$\frac{a-b}{2}$$