Question

ДАЮ СТО БАЛЛОВ!!!!!! РЕШИТЕ СИСТЕМУ УРАВНЕНИЙ ПОЖАЛУЙСТА!!!!​

Answer 1

$\displaystyle\\\left \{ {{x^2-2xy-y^2=2} \atop {xy+y^2=4}} \right.\Leftrightarrow\left \{ {{x^2-2xy-y^2=2} \atop {x=\dfrac{4-y^2}{y} }} \right.$

подставим значение х в первое уравнение

$\displaystyle\\\frac{(4-y)^2}{y^2}-2y\cdot\frac{4-y^2}{y} -y^2=2\\\\\\\frac{16-8y^2+y^4}{y^2} -8+2y^2-y^2=2\\\\\frac{16-8y^2+y^4}{y^2} =10-y^2\\\\16-8y^2+y^4=10y^2-y^4\\\\2y^4-18y^2+16=0\\\\y^4-9y^2+8=0\\\\po~Vieta~y^2=1~ili~y^2=8\\\\(y^2-1)(y^2-8)=0\\\\(y-1)(y+1)(y-2\sqrt{2})(y+2\sqrt{2} )=0\\\\1)y=1;x=(4-1)/1=3\\\\2)y=-1;x=(4-1)/(-1)=-3\\\\3)y=2\sqrt{2} ;x=\frac{4-8}{2\sqrt{2} } =\frac{-2}{\sqrt{2} } =-\sqrt{2} \\\\4)y=-2\sqrt{2} ;x=\frac{4-8}{-2\sqrt{2} } =\sqrt{2}$

$\boldsymbol{Otvet:(-3;-1)~(3;1)~(-\sqrt{2}; 2\sqrt{2} )~(\sqrt{2}; -2\sqrt{2} )}$