Respuesta :

[tex]\frac{\frac{3}{a^2-81}}{\frac{9}{a-9}}[/tex]

the division of fractions follows the stes

[tex]\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a\cdot d}{b\cdot c}[/tex]

Applying this into the expression given

[tex]\frac{3\cdot(a-9)}{9\cdot(a^2-81)}[/tex]

simplify the coefficients by decomposing 9 into a product and cancelling the common factors

[tex]\frac{3\cdot(a-9)}{3\cdot3\cdot(a^2-81)}[/tex]

simplify

[tex]\frac{a-9}{3\cdot(a^2-81)}[/tex]

In the denominators there is a difference of squares that can be rewriten as a product, in which by definition the difference of square is described as

[tex]a^2-b^2=(a+b)\cdot(a-b)[/tex]

In the denominator of the expression we can see that a is squared and that 81 has an exact root which is 9, reason why we can write this as a difference of squares, it should look like this:

simplify the expression

[tex]\frac{1}{3\cdot(a+9)}[/tex]

distribute the 3

[tex]\frac{1}{3a+27}[/tex]