Question

Даю 50 баллов. Решите систему уравнений.

Answer 1

Ответ:

$(x,\ y) = \{(2\sqrt{6},\ 2\sqrt{2});\ (2\sqrt{6},\ -2\sqrt{2})\}$

Объяснение:

$1) \ x-y+\sqrt{\frac{x-y}{x+y} } = \frac{20}{x+y} \ \ \ \ |*(x+y) \neq 0\\(x+y)(x-y+\sqrt{\frac{x-y}{x+y}}) = 20\\(x+y)(x-y) + (x+y)\sqrt{\frac{x-y}{x+y}} = 20\\(x^2-y^2) + \sqrt{\frac{(x+y)^2(x-y)}{x+y}} = 20\\(x^2-y^2) + \sqrt{(x+y)(x-y)} = 20\\(x^2-y^2) + \sqrt{(x^2-y^2)} = 20\\\\t = \sqrt{(x^2-y^2)}, \ t \geq 0\\t^2 + t = 20\\t^2 +t-20 = 0\\D = 1-4*1*(-20) = 81 = 9^2\\t_1 = \frac{-1+9}{2} = 4, t_2 = \frac{-1-9}{2} = - 5\\t_1 \geq 0, \ t_1 \in OD3\\t_2 < 0,\ t_2 \notin OD3$

$\sqrt{(x^2-y^2)} = 4\\x^2-y^2 = 4^2\\x^2-y^2 = 16$

$\left \{ {{x^2-y^2 = 16\ \ \ \ \ \ (1)} \atop {x^2+y^2=32\ \ \ \ \ \ (2)}} \right. \\\\(1)\ +\ (2):\\x^2+x^2-y^2+y^2 = 16+32\\2x^2 = 48\\x^2 = 24\\x = б\sqrt{24} = б2\sqrt{6} \\\\x^2 = 24\ in\ (2):\\24+y^2 = 32\\y^2 = 32-24\\y^2 = 8\\y = б\sqrt{8} = б2\sqrt{2} \\\\(x,\ y) = \{(2\sqrt{6},\ 2\sqrt{2});\ (2\sqrt{6},\ -2\sqrt{2});\ (-2\sqrt{6},\ 2\sqrt{2});\ (-2\sqrt{6},\ -2\sqrt{2})\}$

Проверка:

$x-y+\sqrt{\frac{x-y}{x+y} } = \frac{20}{x+y}\\\\(2\sqrt{6},\ 2\sqrt{2}):\\\\\frac{5\sqrt{6}-5\sqrt{2}}{2} = \frac{5\sqrt{6}-5\sqrt{2}}{2}\ \ (true)\\\\(2\sqrt{6},\ -2\sqrt{2}):\\\\\frac{5\sqrt{6}+5\sqrt{2}}{2} = \frac{5\sqrt{6}+5\sqrt{2}}{2}\ \ (true)\\\\(-2\sqrt{6},\ 2\sqrt{2}):\\\\-2\sqrt{6}-2\sqrt{2}+\sqrt{\sqrt{3} + 2 } = -\frac{5\sqrt{6}+5\sqrt{2}}{2}\ \ (false)\\\\(-2\sqrt{6},\ -2\sqrt{2}):\\\\-2\sqrt{6}+2\sqrt{2}+\sqrt{2-\sqrt{3}} = \frac{-5\sqrt{6}+5\sqrt{2}}{2}\ \ (false)$

$(x,\ y) = \{(2\sqrt{6},\ 2\sqrt{2});\ (2\sqrt{6},\ -2\sqrt{2})\}$