Question

2x^2+4xy-5y=1
x^2+xy-6y^2=0
Эти два уравнения в системе

Answer 1

$\left \{ {{2x^2+4xy-5y^2=1} \atop {x^2+xy-6y^2=0}} \right. \\\\x^2+xy-6y^2=0\; |:y^2\ne 0\\\\(\frac{x}{y})^2+\frac{x}{y}-6=0\\\\t=\frac{x}{y}\; ,\; \; t^2+t-6=0\; \; ,\; \; t_1=2\; ,\; t_2=-3\; \; (teorema\; Vieta)\\\\a)\; \; \frac{x}{y}=2\; \; \to \; \; x=2y\\\\2x^2+4xy-5y^2=2(2y)^2+4\cdot 2y\cdot y-5y^2=8y^2+8y-5y^2=11y^2=1\\\\y=\pm \frac{1}{\sqrt{11}}\; \; \to \; \; \; \; x=\pm \frac{2}{\sqrt{11}}\\\\b)\; \; \frac{x}{y}=-3\; \; \to \; \; x=-3y\\\\2x^2+4xy-5y^2=18y^2-12y^2-5y^2=y^2=1\\\\y=\pm 1\; \; \to \; \; x=\mp3$

$Otvet:\; \; (\frac{2}{\sqrt{11}},\frac{1}{\sqrt{11}} )\; ,\; (-\frac{2}{\sqrt{11}},-\frac{1}{\sqrt{11}})\; ,\; (1,-3)\; ,\; (-1,3)\; .$