Помогите а и б плиз дам 39 баллов

А) $= \frac{a*(a+3)-3(a+3)+18}{a+3} * \frac{ a^{2}+6a+9 }{ a^{2}+9 } * \frac{1}{a-3} = \frac{ a^{2}+3a-3a-9+18 }{a+3} * \frac{ (a+3)^{2} }{ a^{2}+9 } * \frac{1}{a-3} =$ $( a^{2} +9)* \frac{a+3}{ a^{2}+9 } * \frac{1}{a-3} =(a+3)* \frac{1}{a-3} = \frac{a+3}{a-3}$

б) = $( \frac{x}{ y^{2}+xy } + \frac{x-y}{x(x-y)}):( \frac{ y^{2} }{x( x^{2} - y^{2}) } + \frac{1}{x-y} = ( \frac{x}{ y^{2}+xy } + \frac{1}{x}) :( \frac{ y^{2} }{x(x-y)(x+y)} + \frac{1}{x-y} )$ = $\frac{ x^{2} + y^{2}+xy }{x( y^{2}+xy) } : \frac{ y^{2}+x(x+y) }{x(x+y)(x-y)} = \frac{ x^{2} + y^{2}+xy }{xy(y+x)} : \frac{ y^{2}+ x^{2} +xy }{x(x+y)(x-y)} =$ $\frac{x^2+ y^{2}+xy } {xy(y+x)} * \frac{x(x+y)(x-y)}{y^2+x^2+xy}= \frac{1}{y} (x-y)= \frac{x-y}{y}$