1) 5y+1 / y + 1 = y+2 / 2 2) y(в квадрате )/ y + 3= y / y+3
1) 5y+1/y+1=y+2/2 5y+1/y+1=y+2+2, y≠-1 2(5y+1)=(y+2)(y+1) 2(5y+1)-(y+2)(y+1)=0 10y+2-(y²+y+2y+2)=0 10y+2-(y²+3y+2)=0 10y+2-y²-3y-2=0 7y-y=0 y(7-y)=0 y=0 7-y=0 y=0 y=7, y≠-1 y₁=0 y₂=7