Question

поможіть пжжжжжжжжжжжжжж даю 50 баліввв

Answer 1

Ответ:

$\displaystyle \frac{96a^3}{a^4-8a^2+16}$

Объяснение:

$\displaystyle \left(\frac{a-2}{a+2}-\frac{a+2}{a-2}\right)\cdot\frac{12a^2}{4-a^2}=\\\\\\\left(\frac{(a-2)^2}{(a+2)(a-2)}-\frac{(a+2)^2}{(a+2)(a-2)}\right)\cdot\frac{-12a^2}{a^2-4}=\\\\\\\frac{(a-2)^2-(a+2)^2}{(a+2)(a-2)}\cdot\frac{-12a^2}{(a-2)(a+2)}=\\\\\\\frac{a^2-4a+4-(a^2+4a+4)}{(a+2)(a-2)}\cdot\frac{-12a^2}{(a-2)(a+2)}=$

$\displaystyle \frac{a^2-4a+4-a^2-4a-4}{(a+2)(a-2)}\cdot\frac{-12a^2}{(a-2)(a+2)}=\\\\\\\frac{-8a}{(a+2)(a-2)}\cdot\frac{-12a^2}{(a-2)(a+2)}=\\\\\\\frac{96a^3}{((a+2)(a-2))^2}=\\\\\\\frac{-8a}{(a+2)(a-2)}\cdot\frac{-12a^2}{(a-2)(a+2)}=\\\\\\\frac{96a^3}{(a^2-4)^2}=\frac{96a^3}{a^4-8a^2+16}$