Алгебра: Помогите пожалуйста даю 50 пунктов

Решите задачу:

$\frac{b-c}{a+b} - \frac{ab - b^{2} }{ a^{2} -ac}* \frac{a^{2} -c^{2} }{a^{2} -b^{2} } = \\ = \frac{b-c}{a+b} - \frac{b( a - b) }{ a(a-c)}* \frac{(a-c)(a+c)}{(a-b)(a+b)} = \\ = \frac{b-c}{a+b} - \frac{b( a - b)(a-c)(a+c) }{ a(a-c)(a-b)(a+b)}= \\ =\frac{b-c}{a+b} - \frac{b(a+c) }{ a(a+b)}= \frac{a(b-c)}{a(a+b)} - \frac{b(a+c) }{ a(a+b)}=\\ = \frac{ab - ac - ab - bc}{a(a+b)} =\frac{- ac - bc}{a(a+b)} =\frac{- c(a +b)}{a(a+b)} =- \frac{c}{a} \\$

$\frac{ a^{2} -4}{ x^{2} -9}: \frac{ a^{2} -2a}{ xy + 3y}+ \frac{ 2 -y}{ x-3}= \\ = \frac{ (a-2)(a+2)}{ (x-3)(x+3)}: \frac{ a(a-2)}{ y(x+3)}+ \frac{ 2 -y}{ x-3}= \\ = \frac{ (a-2)(a+2)}{ (x-3)(x+3)}* \frac{ y(x+3)}{ a(a-2)}+ \frac{ 2 -y}{ x-3}= \\ = \frac{ (a-2)(a+2) y(x+3)}{ (x-3)(x+3) a(a-2)} + \frac{ 2 -y}{ x-3}= \\ = \frac{ (a+2) y}{ (x-3) a} + \frac{ 2 -y}{ x-3}= \frac{ (a+2) y}{ (x-3) a} + \frac{a( 2 -y)}{ (x-3) a}= \\ =\frac{ ay+2y+2a-ay}{ (x-3) a}=\frac{2(a+y)}{ a(x-3) } \\$