Question

Допоможіть будь ласка

Answer 1

Объяснение:

$\left \{ {\sqrt{x} +\sqrt{y} =5} \atop {x+y+4\sqrt{xy}=37 }} \right. .$

Пусть √x=t    √y=v      ⇒

$\left \{ {{t+v=5} \atop {t^2+v^2+4tv}=37} \right.\ \ \ \ \left \{ {{t+v=5} \atop {t^2+2tv+v^2+2tv=37}} \right. \ \ \ \ \left \{ {{t+v=5} \atop {(t+v)^2+2tv=37}} \right. \ \ \ \ \left \{ {{y=t+v=5} \atop {5^2+2tv=37}} \right.\ \ \ \ \\\left \{ {{v=5-t} \atop {25+2tv=37}} \right. \ \ \ \ \left \{ {{v=5-t} \atop {2tv=12\ |:2}} \right. \ \ \ \ \left \{ {{v=5-t} \atop {tv=6}} \right.\ \ \ \ \left \{ {{v=5-t} \atop {t*(5-t)=6}} \right. \ \ \ \ \left \{ {{v=5-t} \atop {5t-t^2=6}} \right.$

$\left \{ {{v=5-t} \atop {t^2-5t+6=0}} \right. \ \ \ \ \left \{ {{v=5-t} \atop {D=1\ \ \sqrt{D}=1 }} \right. \ \ \ \ \left \{ {{v_1=3\ \ \ \ v_2=2} \atop {t_1=2\ \ \ \ t_2=3}} \right. .$

$\left \{ {{\sqrt{y} =3} \atop {\sqrt{x}=2 }} \right. \ \ \ \ \left \{ {{(\sqrt{y})^2 =3^2} \atop {(\sqrt{x})^2=2^2 }} \right.\ \ \ \ \left \{ {{y_1=9} \atop {x_1=4}} \right. .$

$\left \{ {{\sqrt{y} =2} \atop {\sqrt{x}=3 }} \right. \ \ \ \ \left \{ {{(\sqrt{y})^2 =2^2} \atop {(\sqrt{x})^2=3^2 }} \right.\ \ \ \ \left \{ {{y_2=4} \atop {x_2=9}} \right. .$

Ответ: (4;9),  (9;4)