Question

20 отдам нужно очень.

Answer 1

$( \frac{a+1}{2a-2}+ \frac{6}{2a^2-2}- \frac{a+3}{2a-2})* \frac{4a^2-4}{3}=( \frac{a+1}{2(a-1)}- \frac{a+3}{2(a+1)}+ \frac{6}{2(a^2-1)})* \frac{4(a^2-1)}{3} \\ \\ \frac{(a+1)^2-(a+3)(a-1)+6}{2(a^2-1)}* \frac{4(a^2-1)}{3}= \\ \\ \frac{(a^2+2a+1-a^2+2a-3+6)*4}{6}= \frac{8(a+1)}{3}$