Question

100 баллов! срочно!
решить систему уравнений способом подстановки
$xy = 1$
${x}^{2} + {y}^{2} = 4$
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Answer 1

$\displaystyle\left \{ {{xy=1} \atop {x^2+y^2=4}} \right. ;\left \{ {{y=\dfrac{1}{x} } \atop {x^2+\dfrac{1}{x^2} =4}} \right.$

$x^4-4x^2+1=0\\D=b^2-4ac=16-4=12\\ x^2=(4\pm2\sqrt{3} )/2=2\pm\sqrt{3} \\x=\pm\sqrt{2\pm\sqrt{3} }$

$x_{1}=-\sqrt{2-\sqrt{3} } ;~~y_1=-\dfrac{1}{\sqrt{2-\sqrt{3} }} =-\sqrt{2+\sqrt{3} } \\\\x_{2}=\sqrt{2-\sqrt{3} } ;~~y_2=\dfrac{1}{\sqrt{2-\sqrt{3} }} =\sqrt{2+\sqrt{3} } \\\\x_{3}=-\sqrt{2+\sqrt{3} } ;~~y_3=-\dfrac{1}{\sqrt{2+\sqrt{3} }} =-\sqrt{2-\sqrt{3} } \\\\x_{4}=\sqrt{2+\sqrt{3} } ;~~y_4=\dfrac{1}{\sqrt{2+\sqrt{3} }} =\sqrt{2-\sqrt{3} }$