Question

Решите систему уравнений! Срочно!

Answer 1

$\left \{ {{\sqrt{xy}\cdot (\sqrt{x}+\sqrt{y})=6} \atop {(\sqrt{x}+\sqrt{y})(x-\sqrt{xy}+y)=9}} \right. \; ,\; \; u=\sqrt{x}+\sqrt{y}\; ,\; \; v= \sqrt{xy}\\\\ODZ:\; \; x\geq 0\; ,\; y\geq 0\\\\x+y=(\sqrt{x}-\sqrt{y})^2-2\sqrt{xy}\\\\ \left \{ {{uv=6} \atop {u(u^2-3v)=9}} \right. \; \left \{ {{v=\frac{6}{u}} \atop {u(u^2-3\cdot \frac{6}{u})=9}} \right. \; \left \{ {{v=\frac{6}{u}} \atop {u^3=27}} \right. \; \left \{ {{v=2} \atop {u=3}} \right. \; \left \{ {{\sqrt{xy}=2} \atop {\sqrt{x}+\sqrt{y}=3}} \right. \\\\t=\sqrt{x}\; ,\; p=\sqrt{y}$

$\left \{ {{tp=2} \atop {t+p=3}} \right. \; \left \{ {{t(3-t)=2} \atop {p=3-t}} \right. \; \left \{ {{t^2-3t+2=0} \atop {p=3-t}} \right.\; \left \{ {{t_1=1\; ,\; t_2=2} \atop {p_1=2\; ,\; p_2=1}} \right. \\\\ \left \{ {{ \sqrt{x}=1} \atop { \sqrt{y}=2}} \right. \; \; \; ili\; \; \; \left \{ {{\sqrt{x}=2} \atop {\sqrt{y}=1}} \right. \\\\\left \{ {{x=1} \atop {y=4}} \right. \; \; \; ili\; \; \; \left \{ {{x=4} \atop {y=1}} \right.\\\\Otvet:\; \; (1,4)\; ,\; (4,1)\; .$