\( |x|=3↔x=3\quad or\quad x=-3\\|y|=\dfrac{1}{3}↔y=\dfrac{1}{3}\quad or\quad y=-\dfrac{1}{3}\\→(x,y)=\{ (3;\dfrac{1}{3});(3;-\dfrac{1}{3});(-3;\dfrac{1}{3});(-3;-\dfrac{1}{3})\\ (*) (x,y)=(3;\dfrac{1}{3})\\→B=3^2-3.3.\dfrac{1}{3}+(\dfrac{1}{3})^2=\dfrac{55}{9}\\(*) (x,y)=(3;-\dfrac{1}{3})\\→B=3^2-3.3.(-\dfrac{1}{3})+(-\dfrac{1}{3})^2=\dfrac{109}{9}\\(*) (x,y)=(-3;\dfrac{1}{3})\\→B=(-3)^2-3.(-3).\dfrac{1}{3}+(\dfrac{1}{3})^2=\dfrac{109}{9}\\(*) (x,y)=(-3;-\dfrac{1}{3})\\→B=(-3)^2-3.(-3).(-\dfrac{1}{3}).(-\dfrac{1}{3})^2=\dfrac{55}{9}\)
