Question

решите систему. Пожалуйста срочно нужно

Answer 1

Сделаем замену :

$x\sqrt{y}=a\\y\sqrt{x}=b\\\\\left \{ {{a+b=6} \atop {a^{2}+b^{2}=20}} \right.\\\\\left \{ {{a=6-b} \atop {(6-b)^{2}+b^{2}=20}} \right.\\\\\left \{ {{a=6-b} \atop {36-12b+b^{2}+b^{2}=20}} \right.\\\\\left \{ {{a=6-b} \atop {2b^{2}-12b+16=0}} \right. \\\\\left \{ {{a=6-b} \atop {b^{2}-6b+8=0}} \right.$

$1)\left \{ {{b_{1} =4} \atop {a_{1}=6-4=2 }} \right.\\\\2)\left \{ {{b_{2} =2} \atop {a_{2}=6-2=4 }} \right.\\\\1):\left \{ {{y\sqrt{x} =4} \atop {x\sqrt{y}=2 }} \right.\\ -----\\ \frac{\sqrt{y} }{\sqrt{x} }=2\\ \sqrt{y}=2\sqrt{x}\\ x*2\sqrt{x} +4x\sqrt{x}=6\\2x\sqrt{x}+4x\sqrt{x}=6\\6x\sqrt{x}=6\\x\sqrt{x}=1\\(\sqrt{x})^{3}=1\\x=1\\\sqrt{y}=2\sqrt{1}\\\sqrt{y}=2\\y=4$

$2):\left \{ {{y\sqrt{x} =2} \atop {x\sqrt{y} =4}} \right.\\ -------\\\frac{\sqrt{y} }{\sqrt{x} }=\frac{1}{2}\\\sqrt{x}=2\sqrt{y}\\(2\sqrt{y})^{2}*\sqrt{y}+y*2\sqrt{y}=6\\\\4y\sqrt{y}+2y\sqrt{y}=6\\6y\sqrt{y}=6\\y\sqrt{y}=1\\(\sqrt{y})^{3}=1\\y=1\\\sqrt{x}=2\sqrt{1}\\\sqrt{x}=2\\x=4\\\\Otvet:(1;4),(4;1)$