Алгебра. 7 класс. 14 баллов. Упростить выражение.

Question 1

Question 2

Надо сделать 1 и 2.

Question 3

1)
$( \frac{b}{a^2-ab}+ \frac{a}{b^2-ab})* \frac{a^2b+ab^2}{a^2-b^2}=( \frac{b}{a(a-b)}+ \frac{a}{-b(a-b)})* \frac{a^2b+ab^2}{a^2-b^2}=$ $=( \frac{b}{a(a-b)}- \frac{a}{b(a-b)})* \frac{a^2b+ab^2}{a^2-b^2}= \frac{b^2-a^2}{ab(a-b)}* \frac{ab(a+b)}{(a-b)(a+b)}=$ $= \frac{-(a^2-b^2)}{(a-b)}* \frac{1}{a-b}= - \frac{(a-b)(a+b)}{(a-b)^2} =- \frac{a+b}{a-b}= \frac{a+b}{b-a}$
2)
$( \frac{2a}{a+2}+ \frac{2a}{6-3a} + \frac{8a}{a^2-4}): \frac{a-4}{a-2}=( \frac{2a}{a+2}+ \frac{2a}{3(2-a)} + \frac{8a}{(a-2)(a+2)}): \frac{a-4}{a-2} =$ $=( \frac{2a}{a+2}- \frac{2a}{3(a-2)} + \frac{8a}{(a-2)(a+2)})* \frac{a-2}{a-4} =$ $=\frac{6a(a-2)-2a(a+2)+8a*3}{3(a+2)(a-2)}* \frac{a-2}{a-4} =\frac{6a^2-12a-2a^2-4a+24a}{3(a+2)}* \frac{1}{a-4} = \frac{4a^2+8a}{3(a+2)(a-4)}=$ $= \frac{4a(a+2)}{3(a+2)(a-4)}= \frac{4a}{3(a-4)} = \frac{4a}{3a-12}$
3)
$( \frac{a^2+b^2}{a}+b):(( \frac{1}{a^2}+ \frac{1}{b^2})* \frac{a^3-b^3}{a^2+b^2})= \frac{a^2+b^2+ab}{a}:( \frac{b^2+a^2}{a^2b^2}* \frac{a^3-b^3}{a^2+b^2})=$ $=\frac{a^2+b^2+ab}{a}:( \frac{1}{a^2b^2}* \frac{a^3-b^3}{1})=\frac{a^2+b^2+ab}{a}: \frac{a^3-b^3}{a^2b^2}=\frac{a^2+b^2+ab}{a}* \frac{a^2b^2}{a^3-b^3}=$ $=\frac{a^2+b^2+ab}{a}* \frac{a^2b^2}{(a-b)(a^2+b^2+ab)}= \frac{ab^2}{a-b}$
4)
$( \frac{1}{x} + \frac{1}{y}): \frac{x^2+2xy+y^2}{2xy} = \frac{y+x}{xy} * \frac{2xy}{(x+y)^2}= \frac{2}{x+y}$