[tex]\displaystyle\bf\\\left \{ {{\dfrac{x}{y} +\dfrac{y}{x} =\dfrac{10}{3} } \atop {x^{2} -y^{2} =8}} \right.[/tex]

Решим первое уравнение методом замены :

[tex]\displaystyle\bf\\\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 \ \ , \ \ m\neq 0\\\\\\3m^{2} -10m+3=0\\\\D=(-10)^{2} -4\cdot 3\cdot 3=100-36=64=8^{2} \\\\\\m_{1} =\frac{10-8}{6} =\frac{1}{3} \\\\\\m_{2} =\frac{10+8}{6} =3[/tex]

[tex]\displaystyle\bf\\1)\\\\\frac{x}{y} =\frac{1}{3} \\\\y=3x\\\\x^{2} -y^{2} =8\\\\x^{2} -(3x)^{2} =8\\\\x^{2} -9x^{2} =8\\\\-8x^{2} =8\\\\x^{2} =-1\\\\x\in \ \oslash\\\\\\2)\\\\\frac{x}{y} =3\\\\x=3y\\\\x^{2} -y^{2} =8\\\\(3y)^{2} -y^{2} =8\\\\9y^{2} -y^{2} =8\\\\8y^{2} =8\\\\y^{2} =1\\\\y_{1,2} =\pm 1[/tex]

[tex]\displaystyle\bf\\y=-1 \ \ \ \Rightarrow \ \ \ x=3y=3\cdot(-1)=-3\\\\y=1 \ \ \ \Rightarrow \ \ \ x=3y=3\cdot 1=3\\\\\\Otvet \ : \ \Big(-3 \ ; \ -1\Big) \ ; \ \Big(3 \ ; \ 1\Big)[/tex]

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