Question

Решите систему уравнений

Answer 1

$left { {{sqrt{frac{x}{y}}+sqrt{frac{y}{x}}=frac{5}{2}} atop {x+y=10}} right. ; ,; xne 0; ,; yne 0\\t=sqrt{frac{x}{y}}; ,; left { {{t+frac{1}{t}=frac{5}{2}} atop {x+y=10}} right. ; ,; left { {{2t^2-5t+2=0,; tne 0} atop {x+y=10}} right. ; ,; left { {{t=frac{1}{2},; t=2} atop {x+y=10}} right. \\sqrt{frac{x}{y}}=frac{1}{2}; ,; frac{x}{y}=frac{1}{4}; ,; y=4x\\sqrt{frac{x}{y}}=2; ,; frac{x}{y}=4; ,; x=4y$

$a); ; left { {{y=4x} atop {x+y=10}} right. ; ,; left { {{y=4x} atop {5x=10}} right. ; ,; left { {{y=8} atop {x=2}} right. \\b); ; left { {{x=4y} atop {x+y=10}} right. ; ,; left { {{x=4y} atop {5y=2}} right. ; ,; left { {{x=8} atop {y=2}} right. \\Otvet:; ; (2,8); ,; (8,2); .$

Answer 2

ОДЗ
x/y>0⇒
{x>0            { x<0
{y>0 или     {Y<0
√x/y=a⇒√y/x=1/a
a+1/a=5/2
2a²-5a+2=0
D=25-16=9
a1=(5-3)/4=1/2⇒√x/y=1/2⇒x/y=1/4⇒y=4x
  x+y=10
x+4x=10
5x=10
x=2
y=8
a2=(5+3)/4=2⇒√x/4=2⇒x/y=4 ⇒x=4y
4y+y=10
5y=10
y=2
x=8
Ответ (2;8);(8;2)