Question

Помогите решить пару примеров задание очень трудное полное решение

Answer 1

Немного корявый ответ

Answer 2

266.
1)
$\frac{a^2-ab}{a^2b-b^3}+\frac{2a^2}{b^2-ab^2+a^2b-a^3}-\frac{a}{a^2-2ab+b^2}=\\ =\frac{a(a-b)}{b(a-b)(a+b)}-\frac{a}{(a+b)^2}+\frac{2a^2}{b^2-ab^2+a^2b-a^3}=\\ =\frac{a}{b(a+b)}-\frac{a}{(a+b)^2}+\frac{2a^2}{b^2-ab^2+a^2b-a^3}=\\ =\frac{a^2+ab-ab}{b(a+b)^2}+...=\\ =\frac{a^2}{b(a+b)^2}+\frac{2a^2}{b^2-ab^2+a^2b-a^3}=\\ =\frac{a^2b^2-a^3b^2+a^4b-a^5+2ba^4+2a^3b^2+2a^2b^3}{(ba^2+2ab^2+b^3)(b^2-ab^2+a^2b-a^3)}=\\ =\frac{-a^5+3a^4b+a^3b^2+2a^2b^3+a^2b^2}{(...)}=\\ =$
2)
$\frac{1}{a^2-ac-ab+bc}+\frac{2}{b^2-ab-bc+ac}+\frac{1}{c^2-ac-bc+ab}=\\ =\frac{1}{a(a-c)-b(a-c)}+\frac{2}{-b(a-b)+c(a-b)}+\frac{1}{c(c-b)-a(c-b)}=\\ =\frac{1}{(a-b)(a-c)}+\frac{2}{(a-b)(c-b)}+\frac{-1}{(c-b)(a-c)}=\\ =\frac{c-b+2a-2c-a+b}{(a-b)(c-b)(a-c)}=\frac{a-c}{(a-b)(c-b)(a-c)}=\\ =\frac{1}{(a-b)(c-b)}=\frac{1}{b^2-ab-bc+ac};$
3)
$\frac{a+2}{a^3-3a^2-4a+12}-\frac{3-a}{a^2-5a+6}+\frac{6}{9-a^2}=\\ |a^2-5a+6=a^2-3a-2a+6=a(a-3)-2(a-3)=\\ =(a-3)(a-2)=(2-a)(3-a);|\\ |a^3-3a^2-4a+12=|a=-2|=a^3+2a^2-5a^2-10a+6a+12=\\ =a^2(a+2)-5a(a+2)+6(a+2)=(a+2)(a^2-5a+6)=\\ =(a+2)(a-2)(a-3)|\\ =\frac{a+2}{(a+2)(a-2)(a-3)}-\frac{3-a}{(a-3)(a-2)}+\frac{6}{(3-a)(3+a)}=\\ =\frac{1}{(a-2)(a-3)}+\frac{1}{a-2}-\frac{6}{(a-3)(a+3)}=\\ = \frac{a+3+a^2-9-6a+12}{(a-2)(a^2-9)}=\frac{a^2-5a+6}{(a-2)(a^2-9)}=\\ =\frac{(a-3)(a-2)}{(a-2)(a-3)(a+3)}=\frac{1}{a+3};$
4)
$\frac{21-7b}{b^3+2b^2-9b-18}+\frac{4-b}{b^2+5b+6}-\frac{2-b}{4-b^2}=\\ |b^2+5b+6=(b+2)(b+3);|\\ |b^3+2b^2-9b-18=b^3-3b^2+5b^2-15b+6b-18=\\ =b^2(b-3)+5b(b-3)+6(b-3)=\\ =(b-3)(b^2+5b+6)=(b-3)(b+3)(b+2);|\\ =\frac{7(3-b)}{(b-3)(b+3)(b+2)}+\frac{4-b}{(b+2)(b+3)}-\frac{2-b}{(2-b)(2+b)}=\\ =\frac{-7}{(b+3)(b+2)}+\frac{4-b}{(b+3)(b+2)}-\frac{1}{b+2}=\\ =\frac{-7+4-b-b-3}{(b+3)(b+2)}=\frac{-2b-6}{(b+3)(b+2)}=\frac{-2(b+3)}{(b+3)(b+2)}=\\ =\frac{-2}{b+2};$