Question

Решите систему уравнений

Answer 1

$\displaystyle\bf\\\left \{ {{\dfrac{x^{2} }{y^{2} } +\dfrac{y^{2} }{x^{2} }=\dfrac{x}{y} +\dfrac{y}{x} } \atop {x^{2} +y^{2} =2}} \right. \\\\\\\frac{x}{y} =m \ \ \Rightarrow \ \ \frac{y}{x} =\frac{1}{m}\\\\\\\frac{x^{2} }{y^{2} } =m^{2} \ \ \Rightarrow \ \ \frac{y^{2} }{x^{2} } =\frac{1}{m^{2} } \\\\\\m^{2} +\frac{1}{m^{2} } =m+\frac{1}{m} \\\\\\\Big(m+\frac{1}{m} \Big)^{2} -2=m+\frac{1}{m} \\\\\\m+\frac{1}{m} =t\\\\\\t^{2} -t-2=0\\\\\\t_{1} =-1\\\\t_{2} =2$

$\displaystyle\bf\\1) \ \ m+\frac{1}{m} =-1\\\\\\m+\frac{1}{m} +1=0\\\\\\\frac{m^{2}+m+1 }{m} =0\\\\\\m^{2} +m+1=0 \ , \ m\neq 0\\\\D=1^{2} -4\cdot 1=1-4=-3<0\\\\m\in\oslash\\\\2) \ \ m+\frac{1}{m} =2\\\\\\m+\frac{1}{m} -2=0\\\\\\\frac{m^{2} -2m+1}{m} =0\\\\m^{2}-2m+1=0 \ , \ m\neq 0\\\\(m-1)^{2} =0\\\\m-1=0\\\\m=1\\\\\\\left \{ {{\dfrac{x}{y} =1} \atop {x^{2} }+y^{2} =2} \right. \\\\\\\left \{ {{x=y} \atop {x^{2} +x^{2} =2}} \right.$

$\displaystyle\bf\\\left \{ {{x=y} \atop {2x^{2} =2}} \right. \\\\\\\left \{ {{x=y} \atop {x^{2} =1}} \right. \\\\\\\left[\begin{array}{ccc}\left \{ {{x_{1} =-1} \atop {y_{1} =-1}} \right. \\\left \{ {{x_{2} =1} \atop {y_{2} =1}} \right. \end{array}\right\\\\\\Otvet:(-1 \ ; \ -1) \ , \ (1 \ ; \ 1)$

Answer 2

Объяснение:

$\left\{\begin{array}{ccc}\frac{x^2}{y^2}+\frac{y^2}{x^2}=\frac{x}{y} +\frac{y}{x} \\x^2+y^2=2\\\end{array}\right\ \ \ \ \left\{\begin{array}{ccc}\frac{x^4+y^4}{x^2y^2}=\frac{x^2+y^2}{xy} \\x^2+y^2=2\\\end{array}\right\ \ \ \ \left\{\begin{array}{ccc}\frac{x^4+y^4}{x^2y^2}=\frac{2}{xy} \\x^2+y^2=2\\\end{array}\right\ \ \ \ x\neq 0,\ \ \ \ y\neq 0.\\x^4+y^4=2*xy\\x^4+2x^2y^2+y^4=2xy+2x^2y^2\\(x^2+y^2)^2=2xy+2x^2y^2\\2^2=2xy+2x^2y^2\\2x^2y^2+2xy-4=0\ |:2\\x^2y^2+xy-2=0$

Пусть ху=$t$       ⇒

$t^2+t-2=0\\D=9\ \ \ \ \ \sqrt{D}=3\\t_1=xy=-2.\ \ \ \ \ \Rightarrow\\\left \{ {{x^2+y^2=2} \atop {xy=-2}} \right. \ \ \ \ \ \left \{ {{x^2+2xy+y^2=2+2xy} \atop {xy=-2}} \right.\ \ \ \ \ \left \{ {(x+y)^2=2+2*(-2)} \atop {xy=--2}} \right.\ \ \ \ \ \left \{ {{(x+y)^2=-2} \atop {xy=-2}} \right. \ \ x, \ y\in \varnothing.\\t_2=xy=1.\ \ \ \ \ \Rightarrow$

$\left \{ {x^2+y^2=2} \atop {xy=1}} \right. \ \ \ \ \ \left \{ {{x^2+2xy+y^2=2+2xy} \atop {xy=1}} \right. \ \ \ \ \ \left \{ (x+y)^2=2+2*1} \atop {xy=1}} \right. \ \ \ \ \left \{ {{(x+y)^2=4} \atop {xy=1}} \right.\\1)\ \left \{ {{x+y=2} \atop {xy=1}} \right. \ \ \ \ \left \{ {{y=2-x} \atop {x*(2-x)=1}} \right. \ \ \ \ \ \left \{ {{y=2-x} \atop {2x-x^2=1}} \right.\ \ \ \ \ \left \{ {{y=2-x} \atop {x^2-2x+1=0}} \right.\ \ \ \ \left \{ {{y=2-x} \atop {(x-1)^2=0}} \right.$

$\left \{ {{y=2-x} \atop {x-1=0}} \right.\ \ \ \ \ \left \{ {{y_1=1} \atop {x_1=1}} \right..$

$2)\ \left \{ {{x+y=-2} \atop {xy=1}} \right. \ \ \ \ \left \{ {{y=-2-x} \atop {x*(-2-x)=1}} \right. \ \ \ \ \ \left \{ {{y=-2-x} \atop {-2x-x^2=1}} \right.\ \ \ \ \ \left \{ {{y=-2-x} \atop {x^2+2x+1=0}} \right.\ \ \ \ \left \{ {{y=-2-x} \atop {(x+1)^2=0}} \right.$

$\left \{ {{y=-2-x} \atop {x+1=0}} \right.\ \ \ \ \ \left \{ {{y_2=-1} \atop {x_2=-1}} \right..$

Ответ: (1;1),  (-1;-1).