Question

решите систему плиз
x^2+y^2=10
xy=3

Answer 1

$left { {{x^2+y^2=10} atop {xy=3}} right. \x= frac{3}{y} \(frac{3}{y})^2+y^2=10 \ frac{9}{y^2} +y^2=10 \y neq 0 \9+y^4=10y^2 \y^4-10y^2+9=0 \t=y^2, t textgreater 0 \t^2-10t+9=0 \D=100-36=64=8^2 \t_1= frac{10+8}{2} =9 \t_2=1 \y^2=9 \y_1=3 \y_2=-3 \y^2=1 \y_3=1 \y_4=-1 \x_1= frac{3}{3}=1 \x_2=-1 \x_3= frac{3}{1}=3 \x_4=-3$
Ответ: (1;3), (3;1), (-3;-1), (-1;-3)