Question

51 номер пожалуйста

Answer 1

$\displaystyle\bf\\\frac{xy}{x^{2} +y^{2} } =0,3\\\\\\\frac{xy}{x^{2} +y^{2} } =\frac{3}{10} \\\\\\\frac{x^{2} +y^{2} }{xy} =\frac{10}{3} \\\\\\\frac{x^{2} }{xy} +\frac{y^{2} }{xy} =\frac{10}{3} \\\\\\\frac{x}{y} +\frac{y}{x}=\frac{10}{3} \\\\\\\frac{x}{y} =m \ \ \ \Rightarrow \ \ \ \frac{y}{x} =\frac{1}{m} \\\\\\m+\frac{1}{m} -\frac{10}{3} =0\\\\\\\frac{3m^{2} -10m+3}{3m} =0\\\\\\3m^{2} -10m+3=0 \ \ , \ \ m\neq 0\\\\\\D=(-10)^{2} -4\cdot 3\cdot 3=100-36=64=8$

$\displaystyle\bf\\m_{1} =\frac{10-8}{6} =\frac{1}{3} \\\\\\m_{2}=\frac{10+8}{6} } =3\\\\\\ili \ \ \frac{x}{y}=\frac{1}{3} \ \ \ ; \ \ \ ili \ \ \frac{x}{y} =3$

Так как надо найти наименьшее значение , то ответ : 1\3