Question

$( \frac{a+3}{a-3} + \frac{a-3}{a+3}) : \frac{3a^2+27}{9-a^{2} }$Упростите выражение (a+3/a−3+a−3/a+3):3a^2+27/9−a^2.

Answer 1

$(\frac{a+3}{a-3}+\frac{a-3}{a+3}):\frac{3a^2+27}{9-a^{2}}=$

$(\frac{(a+3)^2}{(a-3)(a+3)}+\frac{(a-3)^2}{(a+3)(a-3)})*\frac{9-a^{2}}{3a^2+27}=$

$\frac{a^2+6a+9+a^2-6a+9}{a^2-9}*\frac{-(a^{2}-9)}{3a^2+27}=$

$\frac{-2(a^2+9)}{3(a^2+9)}=-\frac{2}{3}$