РЕШИТЕ ПЖ,очень срочно:сс 1)b\a-b:(a\a-b-a+b\a) 2)(a\a-b+a\b):a\a-b...

$\frac{b}{a-b}:( \frac{a}{a-b}- \frac{a+b}{a})= \frac{b}{a-b}: \frac{a^{2}-(a+b)(a-b)}{a(a-b)}= \frac{b}{a-b}* \frac{a(a-b)}{ a^{2}-a^{2}+b^{2}}= \frac{a}{b} \\ \\ (\frac{a}{a-b} + \frac{a}{b}) : \frac{a}{a-b}= \frac{ab+a(a-b)}{b(a-b)}* \frac{a-b}{a} = \frac{ab+ a^{2}-ab }{b} * \frac{1}{a} = \frac{a}{b} \\ \\ (\frac{1}{ b^{2} }- \frac{1}{ a^{2} } )* \frac{ab}{a+b} = \frac{a^{2} - b^{2} }{a^{2} b^{2} } * \frac{ab}{a+b}= \frac{(a-b)(a+b)}{ab(a+b)}= \frac{a-b}{ab} \\ \\4) [tex]\frac{c-3}{c} - \frac{c^{2}-9 }{c} + \frac{3}{c}= \\ \\ \\ \\ \frac{b-2a}{a-b} * \frac{ a^{2}- b^{2} }{4a}= \frac{(b-2a)(a-b)(a+b)}{4a(a-b)}= \frac{(b-2a)(a+b)}{4a}$ [/tex]