Помогите с заданием Докажите тождество заранее спасибо

1) [(a-b)/(a²+ab)-a/(ab+b²)]:[1/(a+b)-b²/(ab²-a³)]= =[(a-b)/{a(a+b)}-a/{b(a+b)}]:[1/(a+b)-b²/{a(b²-a²)}]= =[{(a-b)b-a²}/{ab(a+b)}]:[{a(b-a)-b²}/{a(b²-a²)}]= = [(ab-b²-a²)/{ab(a+b)}]:[(ab-b²-a²)/{a(b²-a²)}]=(b-a)/b ч.т.д. 2) {2ab/(a²-b²)+(a-b)/[2(a+b)]}·2·a/(a+b)-b/(a-b)=1 {2ab/(a²-b²)+(a-b)/[2(a+b)]}·2·a/(a+b)-b/(a-b)= =[{2·2ab+(a-b)²}/(2(a²-b²))]·[2a/(a+b)]-b/(a-b)= =[{4ab+(a²-2ab+b²)}/(2(a²-b²))]·[2a/(a+b)]-b/(a-b)= =[{(a²+2ab+b²)}/(2(a²-b²))]·[2a/(a+b)]-b/(a-b)= =[{(a+b)²}/(2(a-b)(a+b))]·[2a/(a+b)]-b/(a-b)= =[(a+b)/(a-b)]·[a/(a+b)]-b/(a-b)=[a/(a-b)]-b/(a-b)=[a-b]/(a-b)=1 ч.т.д. =

Решение в приложении.