Question

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Answer 1

Объяснение:

$\displaystyle\\\left \{ {{\dfrac{x}{y}+\dfrac{y}{x} =\dfrac{5}{2} } \atop {x^2-y^2=27}} \right.\qquad\ x\neq 0;y\neq 0\\\\\\t=\dfrac{x}{y};t+\frac{1}{t} =\frac{5}{2};2t^2-5t+2=0\\\\\\D=25-4\cdot4=9=3^2\\\\\\t_1=\frac{5+3}{4}=2;\frac{x}{y}=2;4y^2-y^2=27;y^2=9;y=\pm3;x=\pm6\\\\\\t_2=\frac{5-3}{4}=\frac{1}{2} ;\frac{x}{y}=\frac{1}{2} ;x^2-4x^2=27;x^2=-9;=\O\\\\\\\\Otvet:\boxed{\bold(-6;-3);(6;3) }$