Question

Simplify the complex rational expression by writing it as division:

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

Short answer

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

Step-by-step solution

Step 1. Given information.

The following complex rational expression is given:

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

Step 2. Simplification of the complex rational expression.

Find the common denominators for the numerator and the denominator:

$\frac{{\displaystyle \frac{1.b}{a.b}+\frac{1.a}{b.a}}}{{\displaystyle \frac{1.b^2}{a^2.b^2}-\frac{1.a^2}{b^2.a^2}}}$

Simplify the numerator and the denominator:

$$\begin{gathered} \frac{{\displaystyle \frac{b}{ab}+\frac{a}{ab}}}{{\displaystyle \frac{b^2}{a^2{b}^2}-\frac{a^2}{a^2{b}^2}}} \\ \frac{{\displaystyle \frac{b+a}{ab}}}{{\displaystyle \frac{b^2-a^2}{a^2{b}^2}}} \end{gathered}$$

Rewrite as a division problem:

$\frac{b+a}{ab}÷\frac{b^2-a^2}{a^2{b}^2}$

Rewrite as the product of the first and the reciprocal of the second:

$\frac{b+a}{ab}.\frac{a^2{b}^2}{b^2-a^2}$

Factorise the terms (where possible), cancel the common factors from the numerator and the denominator and simplify:

localid="1662391087352" $$\begin{gathered} \frac{b+a}{ab}.\frac{ab.ab}{\left(b+a\right)\left(b-a\right)} \\ \frac{ab}{b-a} \end{gathered}$$