Question

СРОЧНО ПОМОГИТЕ!!! ОЧЕНЬ НУЖНО

Answer 1

Ответ:

Формула:   $(x-y)^2=x^2-2xy+y^2\ \ \Rightarrow \ \ \ \ \boxed{\ x^2+y^2=(x-y)^2+2xy\ }$

$\left\{\begin{array}{l}x-y=2\\x^2+y^2=20\end{array}\right\ \ \left\{\begin{array}{l}x-y=2\\(x-y)^2+2xy=20\end{array}\right\ \ \left\{\begin{array}{l}x-y=2\\2^2+2xy=20\end{array}\right$

$\left\{\begin{array}{l}x-y=2\\2xy=16\end{array}\right\ \ \left\{\begin{array}{l}x=y+2\\2xy=16\end{array}\right\ \ \left\{\begin{array}{l}x=y+2\\2(y+2)\, y=16\end{array}\right\ \ \left\{\begin{array}{l}x=y+2\\2y^2+4y-16=0\end{array}\right$

$\left\{\begin{array}{l}x=y+2\\y^2+2y-8=0\end{array}\right\ \ \left\{\begin{array}{l}x_1=-2\ ,\ x_2=4\\y_1=-4\ ,\ y_2=2\end{array}\right\\\\\\y^2+2y-8=0\ \ \to\ \ \ y_1=-4\ ,\ y_2=2\ \ (teorema\ Vieta)\\\\\\Otvet:\ \ (-2;-4)\ ,\ (4;2)\ .$