решить систему равенст
x+y=4
xy=3

x = 4 - y(4 - y)*y = 3 **4y - y^2 - 3 = 0 /* (-1)y^2 - 4y + 3 = 0 D = 16 - 12 = 4y1 = (4 + 2)/2 = 3y2 = (4 - 2)/2 = 1y1 = 3x1 = 4 - y1 = 4 - 3 = 1y2 = 1x2 = 4 - y2 = 4 - 1 = 3Ответ(1; 3 ) ; (3; 1)