Question

Решите систему уравнений {x^2-2xy-y^2=1 {x+y=2

Answer 1

$\left \{ {{x^2-2xy-y^2=1} \atop {x+y=2}} \right. \left \{ {{x^2-2xy-y^2=1} \atop {x=2-y}} \right. \\(2-y)^2-2(2-y)y-y^2=1\\2y^2-8y+3=0\\D_1=16-6=10\\y_1= \frac{4+ \sqrt{10} }{2} \\y_2=\frac{4- \sqrt{10} }{2} \\x_1=2-\frac{4+ \sqrt{10} }{2} \\x_1=- \frac{ \sqrt{10} }{2}\\x_2=2-\frac{4- \sqrt{10} }{2} \\x_2= \frac{ \sqrt{10} }{2} \\........................................................................................\\ (x_1;y_1) =(- \frac{ \sqrt{10} }{2} ;\frac{4+ \sqrt{10} }{2} )\\(x_2;y_2)=<img src=$" alt="( \frac{ \sqrt{10} }{2} ;\frac{4- \sqrt{10} }{2} )" />" />