Question

решите систему уравнений​

Answer 1

Объяснение:

$\left \{ {{x^2-y^2=1} \atop {x^2+xy=2}} \right. \ \ \ \ \ \left \{ {{(x+y)*(x-y)=1} \atop {x*(x+y)=2}} \right. \ \ \ \ \ \left \{ {{\frac{2}{x} *(x-y)=1} \atop {x+y=\frac{2}{x} }} \right.\ \ \ \ \left \{ {{x-y=\frac{x}{2} } \atop {x+y=\frac{2}{x} }} \right. .$

Суммируем эти уравнения:

$2x=\frac{x}{2}+\frac{2}{x} \ |*2x\ \ \ \ \ x\neq 0\\4x^2=x^2+4\\3x^2=4\ |:4\\x^2=\frac{4}{3} \\x_{1,2}=б\sqrt{\frac{4}{3} } =б\frac{2}{\sqrt{3} } =б\frac{2\sqrt{3} }{3} .\\x_1=\frac{2\sqrt{3} }{3}.\\ (\frac{2\sqrt{3} }{3})^2+\frac{2\sqrt{3} }{3}*y=2\\\frac{4}{3} +\frac{2\sqrt{3} }{3}*y =2\\\frac{2\sqrt{3} }{3}*y =\frac{2}{3} \ |*\frac{3}{2}\\\sqrt{3} *y=1 \ |:\sqrt{3} \\y_1=\frac{1}{\sqrt{3} }=\frac{\sqrt{3} }{3}. \\x_2=-\frac{2\sqrt{3} }{3}.$

$(-\frac{2\sqrt{3} }{3})^2+(-\frac{2\sqrt{3} }{3})*y=2\\\frac{4}{3} -\frac{2\sqrt{3} }{3}*y =2\\\frac{2\sqrt{3} }{3}*y =-\frac{2}{3} \ |*\frac{3}{2}\\\sqrt{3} *y=-1 \ |:\sqrt{3} \\y_2=-\frac{1}{\sqrt{3} }=-\frac{\sqrt{3} }{3}.$

Ответ: (2√3/3;√3/3)    (-2√3/3;-√3/3).