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]