Question

СРОЧНО!!! ДАЮ 50 БАЛЛОВ!!!
Решите систему уравнений
$\frac{x}{y} + \frac{y}{x} = \frac{5}{2}$
{
${x}^{2} - {y}^{2} = 27$
​

Answer 1

Ответ:

Объяснение:

$\displaystyle\\\left \{ {{\dfrac{x}{y} +\dfrac{y}{x}=\dfrac{5}{2} } \atop {x^2-y^2=27}} \right. ~~~~~~x\neq 0;y\neq 0\\\\\\\left \{ {{2x^2+2y^2=5xy} \atop {x^2-y^2=27}} \right. ;\left \{ {{2y^2-5xy+2y^2=0} \atop {x^2-y^2=27}} \right. \\\\\\\left \{ {{2y^2-4xy-xy+2x^2=0} \atop {x^2-y^2=27}} \right. ;\left \{ {{2y(y-2x)-x(y-2x)=0} \atop {x^2-y^2=27}} \right. \\\\\\\left \{ {{(y-2x)(2y-x)=0} \atop {x^2-y^2=27}} \right.$

$\displaystyle\\1)\left \{ {{y=2x} \atop {x^2-y^2=27}} \right. ;\left \{ {{y=2x} \atop {x^2-4x^2=27}} \right. ;\left \{ {{y=2x} \atop {x^2=-1}} \right.$   ∅

$\displaystyle\\2)\left \{ {{x=2y} \atop {4y^2-y^2=27}} \right. ;\left \{ {{x=2y} \atop {y^2=9}} \right.;\left \{ {{x=2y} \atop {y=\pm3}} \right. \\\\y_1=-3;x_1=-6\\y_2=3;x_2=6\\\\Otvet:(-6;-3)~(6;3)$