Ответ:
N1)a)30(a²-b²)/20(a+b)=30(a+b)(a-b)/20(a+b)=3(a-b)/2
б)(2a-3b)(2a-3b)/(2a-3b)(2a+3b)=(2a-3b)/(2a+3b)
N2)a)3(x²-2xy+y²)/6(x²-y²)=3(x-y)(x-y)/6(x-y)(x+y)=(x-y)/2(x+y)=(3-(-5))/2(3-5)=(3+5)/2(-2)=8/(-4)=-2
б)(3x(a+b)-y(a+b))/(x(a+b)-3y(a+b))=(a+b)(3x-y)/(a+b)(x-3y)=(3x-y)/(x-3y)