ответ во вложении...
1)(6а/(2а+5))-(16а/(4а²+20а+25)=((6а/(2а+5)²)-(16а/(2а+5)²)=
(12а²+30а-16а)/(2а+5)²= (12а²+14а)/(2а+5)²=2а(6а+7)/((2а+5)²)
2) (2а(6а+7)/(2а+5)²):((6а+7)/(4а²-25))=2а*(4а²-25))/((2а+5)²)=
(2а*(2а-5)(2а+5))/((2а+5)²)=(4а²-10а)/(2а+5)
3)((4а²-10а)/(2а+5))+(10а-25)/(2а+5)=(4а²-10а+10а-25)/(2а+5)=
(2а-5)(2а+5)/(2а+5)=2а-5
предпоследний ответ верный
