Question

Dividing algebraic fractions. http://goo.gl/VkXkib please help me

Answer 1

do you can rewrite these fractions here ?

Answer 2

\(\large\tt \color{black}{\dfrac{8}{9-a^2}\div\dfrac{4a^2-4a-24}{a^2-6a+9}}\)

\(\large\tt \color{black}{=\dfrac{8}{9-a^2}\times\dfrac{a^2-6a+9}{4a^2-4a-24}}\)

\(\large\tt \color{black}{=\dfrac{8}{9-a^2}\times\dfrac{a^2-3a-3a+9}{4(a^2-a-6)}}\)

\(\large\tt \color{black}{=\dfrac{8}{3^2-a^2}\times\dfrac{a(a-3)-3(a-3)}{4(a^2-3a+2a-6)}}\)

\(\large\tt \color{black}{=\dfrac{8}{(3+a)(3-a)}\times\dfrac{(a-3)^2}{4(a(a-3)+2(a-3))}}\)

\(\large\tt \color{black}{=\dfrac{-8}{(3+a)(a-3)}\times\dfrac{(a-3)^2}{4(a-3)(a+2)}}\)

\(\large\tt \color{black}{=\dfrac{-2}{(3+a)}\times\dfrac{1}{(a+2)}}\)

\(\large\tt \color{black}{=\dfrac{-2}{(3+a)(a+2)}}\)

Answer 3

thanks so much

Answer 4

yo welcome

Answer 5

actually thts wrong

Answer 6

O:

Answer 7

can u tell me whats the answer given