Question

50 балів ......................

Answer 1

$sqrt{x + y} + sqrt{x - y} = 10 \ sqrt{ {x}^{2} - {y}^{2} } = 9 \ \ sqrt{x + y} + sqrt{x - y} = 10 \ sqrt{ (x - y)(x + y) } = 9 \ \ a = sqrt{x + y} geqslant 0 \ b = sqrt{x - y} geqslant 0 \ \ a + b = 10 \ ab = 9 \ \ b = 10 - a \ a(10 - a) = 9 \ {a}^{2} - 10a + 9 = 0 \ (a - 9)(a - 1) = 0 \ a_1= 9 = >b_1 =10 - a_1 = 1\ a_2 = 1 = > b_2 =10 - a_2 = 9$
$sqrt{x + y} = 9 \ sqrt{x - y} = 1 \ x + y = 81 \ x - y = 1 \ 2x = 82 = > x_1 = 41 = > y_1 = 40\ \ sqrt{x + y} = 1 \ sqrt{x - y} = 9 \ x + y = 1 \ x - y =81 \ 2x = 82 = > x_2 = 41 = > y_2 = - 40$
Ответ
$\ x_1 = 41 \ y_1 = 40 \ \ x_2 = 41 \ y_2 = - 40$