Question

Решите систему уравнения​

Answer 1

Объяснение:

$\left \{ {{\sqrt{x} +\sqrt{y} =10} \atop {\sqrt[4]{x} +\sqrt[4]{y} =4}} \right. .$

Пусть ⁴√x=t    ⁴√y=v.      ⇒

$\left \{ {{t^2+v^2=10} \atop {t+v=4}} \right.\ \ \ \ \left \{ {{t^2+v^2=10} \atop {v=4-t}} \right.\ \ \ \ \left \{ {{t^2+(4-t)^2=10} \atop {v=4-t}} \right.\ \ \ \ \left \{ {{t^2+16-8t+t^2=10} \atop {v=4-t}} \right.\\\left \{ {{2t^2-8t+6=0\ |;2} \atop {v=4-t}} \right. \ \ \ \ \left \{ {{t^2-4t+3=0} \atop {v=4-t}} \right. \ \ \ \ \left \{ {{D=4\ \ \sqrt{D}=2 } \atop {v=4-t}} \right.\ \ \ \ \left \{ {{t_1=1\ \ t_2=3} \atop {v_1=3\ \ v_2=1}} \right. . \ \ \ \ \ \Rightarrow$

$t_1=\sqrt[4]{x}=1\ \ \ \ x_1=1^4=1.\\t_2=\sqrt[4]{x}=3 \ \ \ \ x_2=3^4=81.\\v_1=\sqrt[4]{y}=3\ \ \ \ y_1=3^4=81.\\v_2=\sqrt[4]{y}=1\ \ \ \ y_2=1^4=1.$

Ответ: (1;81),  (81;1).