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 = 4 y1 = (4 + 2)/2 = 3 y2 = (4 - 2)/2 = 1 y1 = 3 x1 = 4 - y1 = 4 - 3 = 1 y2 = 1 x2 = 4 - y2 = 4 - 1 = 3 Ответ (1; 3 ) ; (3; 1)