\sqrt{x} + \sqrt{y}=4 x+y-3\sqrt{xy}=1

√x + √y = 4
x + y - 3√(xy) = 1

x + y + 2√(xy) = 16
x + y - 3√(xy) = 1

5√(xy) = 15

√(xy) = 3

xy = 9

√x + √y = 4
√x = 3/√y

3/√y + √y = 4

y - 4√y + 3 = 0
D = 16 - 12 = 4
√y = (4 - 2)/2 = 1 => √x = 3/1 = 3
y = 1; x = 9
√y = (4 + 2)/2 = 3 => √x = 3/3 = 1
y = 9; x = 1

ответ: (1; 9) и (9; 1)