Solve (1-a^2/a^2-b^2):(a/a-a/a-b)=

$$( 1 - \frac { a ^ { 2 } } { a ^ { 2 } - b ^ { 2 } } ) : ( \frac { a } { a } - \frac { a } { a - b } ) =$$

Evaluate

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

Short Solution Steps

Divide a by a to get 1.

$$\frac{1-\frac{a^{2}}{a^{2}-b^{2}}}{1-\frac{a}{a-b}}$$

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

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{\left(a+b\right)\left(a-b\right)}{\left(a+b\right)\left(a-b\right)}.

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

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

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

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

$$\frac{\frac{a^{2}-ab+ba-b^{2}-a^{2}}{\left(a+b\right)\left(a-b\right)}}{1-\frac{a}{a-b}}$$

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

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{a-b}{a-b}.

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

Since \frac{a-b}{a-b} and \frac{a}{a-b} have the same denominator, subtract them by subtracting their numerators.

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

Combine like terms in a-b-a.

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

Divide \frac{-b^{2}}{\left(a+b\right)\left(a-b\right)} by \frac{-b}{a-b} by multiplying \frac{-b^{2}}{\left(a+b\right)\left(a-b\right)} by the reciprocal of \frac{-b}{a-b}.

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

Cancel out -b\left(a-b\right) in both numerator and denominator.

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

Expand

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

Short Solution Steps

Divide a by a to get 1.

$$\frac{1-\frac{a^{2}}{a^{2}-b^{2}}}{1-\frac{a}{a-b}}$$

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

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{\left(a+b\right)\left(a-b\right)}{\left(a+b\right)\left(a-b\right)}.

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

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

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

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

$$\frac{\frac{a^{2}-ab+ba-b^{2}-a^{2}}{\left(a+b\right)\left(a-b\right)}}{1-\frac{a}{a-b}}$$

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

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{a-b}{a-b}.

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

Since \frac{a-b}{a-b} and \frac{a}{a-b} have the same denominator, subtract them by subtracting their numerators.

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

Combine like terms in a-b-a.

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

Divide \frac{-b^{2}}{\left(a+b\right)\left(a-b\right)} by \frac{-b}{a-b} by multiplying \frac{-b^{2}}{\left(a+b\right)\left(a-b\right)} by the reciprocal of \frac{-b}{a-b}.

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

Cancel out -b\left(a-b\right) in both numerator and denominator.

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