Question

Помогите пожалуйста!!!

Answer 1

$a)\;\frac{3x^2-6xy+3y^2}{6x^2-6y^2}=\frac{3(x^2-2xy+y^2)}{6(x^2-y^2)}=\frac{(x-y)^2}{2(x-y)(x+y)}=\frac{x-y}{2(x+y)}\\\\x=3,\;y=-5:\\\frac{x-y}{2(x+y)}=\frac{3-(-5)}{2(3+(-5))}=\frac{3+5}{2(3-5)}=\frac8{2(-2)}=\frac8{-4}=-2$

$b)\;\frac{3ax-ay+3bx-by}{ax-3ay+bx-3by}=\frac{a(3x-y)+b(3x-y)}{a(x-3y)+b(x-3y)}=\frac{(a+b)(3x-y)}{(a+b)(x-3y)}=\frac{3x-y}{x-3y}\\\\x=-1,\;y=-2:\\\frac{3x-y}{x-3y}=\frac{3\cdot(-1)-(-2)}{-1-3\cdot(-2)}=\frac{-3+2}{-1+6}=-\frac15$