Question

Решить систему уравнений: x^2-2xy+y^2=4 xy=3

Answer 1

{x^2-2xy+y^2=4

{y=3/x

$x^{2}-\frac{6x}{x}+\frac{3^{2}}{x^{2}}-4=0\\x^{2}-6+\frac{9}{x^{2}}-4=0\\ x^{4}-6x^{2}-4x^{2}+9=0\\x^{4}-10x^{2}+9=0\\\\t=x^{2}\\\\ t^{2}-10t+9=0\\t_{1}+t_{2}=10\\t_{1}\cdot t_{2}=9 \\t_{1}=9\\t_{2}=1\\\\x_{1}=3\\x_{2}=-3\\x_{3}=1\\x_{4}=-1\\\\ y_{1}=\frac{3}{3}=1\\ y_{2}=\frac{-3}{3}=-1\\y_{3}=\frac{1}{3}\\y_{4}=\frac{-1}{3}$

$(3;1), (-3;-1), (1;\frac{1}{3}), (-1;-\frac{1}{3})$

Answer 2

откройте вложения, там написано решение задания