\sqrt{x} + \sqrt{y}=4

x+y-3\sqrt{xy}=1

√x + √y = 4x + y - 3√(xy) = 1x + y + 2√(xy) = 16x + y - 3√(xy) = 15√(xy) = 15√(xy) = 3xy = 9√x + √y = 4√x = 3/√y3/√y + √y = 4y - 4√y + 3 = 0D = 16 - 12 = 4√y = (4 - 2)/2 = 1 => √x = 3/1 = 3y = 1; x = 9√y = (4 + 2)/2 = 3 => √x = 3/3 = 1y = 9; x = 1Ответ: (1; 9) и (9; 1)