Question

Разность квадратов двух чисел равна 80 ,а сумма квадратов равна 208 найдите эти числа

Answer 1

$+\left\{\begin{array}{ccc}x^{2}-y^{2}=80\\x^{2}+y^{2} =208 \end{array}\right\\----------\\2x^{2}=288\\\\x^{2}=144 \\\\x=\pm\sqrt{144}=\pm12 \\\\y^{2} =x^{2}-80=144 -80=64\\\\y=\pm\sqrt{64}=\pm8\\\\Otvet:\boxed{(-12 \ ; \ -8) \ , \ (-12 \ ; \ 8 ) \ , \ (12 \ ; \ -8) \ , \ (12 \ ; \ 8)}$

Answer 2

Ответ ............................................................................................................

$\left\{\begin{array}{l}x^2+y^2=208\\x^2-y^2=80\end{array}\right\ \oplus \ominus \ \left\{\begin{array}{l}2x^2=128\\2y^2=288\end{array}\right\ \ \left\{\begin{array}{l}x^2=64\\y^2=144\end{array}\right\ \ \left\{\begin{array}{l}x=\pm 8\\y=\pm 12\end{array}\right\\\\\\Otvet:\ \ (8;12)\ ,\ (8;-12)\ ,\ (-8;12)\ ,\ (-8;-12)\ .$