Răspuns :
[tex]\it x\circ y = 3x+3y-xy-6 =3x+3y-xy-9+3 = \\ \\ = (3x-9)-(xy-3y )+3 = 3(x-3) -y(x-3) +3 = \\ \\ = (x-3)(3-y) +3 = 3-(x-3)(y-3)\ \ \ \ (*)[/tex]
[tex]\it x\ \textless \ 3 \Rightarrow x-3\ \textless \ 0 \ \ \ \ (1) \\ \\ y\ \textless \ 3 \Rightarrow y-3\ \textless \ 0 \ \ \ \ (2) \\ \\ (1), (2) \Rightarrow (x-3)(y-3)\ \textgreater \ 0 |_{\cdot(-1)} \Rightarrow -(x-3)(y-3)\ \textless \ 0|_{+3} \Rightarrow \\ \\ \Rightarrow 3-(x-3)(y-3) \ \textless \ 3 \Rightarrow 3-(x-3)(y-3) \in (-\infty,\ 3) \ \ \ \ (**) [/tex]
[tex]\it (*),(**) \Rightarrow x\circ y \in(-\infty,\ 3) [/tex]
severifica prin calcul ca:
x°y=-9(1-x/3) (1-y/3)+3
intr-adevar
-9(1-x/3-y/3+xy/9)+3=-9+3x+3y-xy+3=3x+3y-xy-6
Atunci
cand x si y∈(-∞;0)
1-x/3 ∈(0;∞)
1-y/3∈(0;∞)
produsulcelor de mai sus va fi strict pozitiv si, inmultt cu numaruil negativ -9, vom avea
-9(1-x/3) (1-y/3)∈(-∞;0) si
adunand 3 vom avea
-9(1-x/3) (1-y/3)+3∈(-∞;3)
fff grea si foarte
as tricky as that!!!
x°y=-9(1-x/3) (1-y/3)+3
intr-adevar
-9(1-x/3-y/3+xy/9)+3=-9+3x+3y-xy+3=3x+3y-xy-6
Atunci
cand x si y∈(-∞;0)
1-x/3 ∈(0;∞)
1-y/3∈(0;∞)
produsulcelor de mai sus va fi strict pozitiv si, inmultt cu numaruil negativ -9, vom avea
-9(1-x/3) (1-y/3)∈(-∞;0) si
adunand 3 vom avea
-9(1-x/3) (1-y/3)+3∈(-∞;3)
fff grea si foarte
as tricky as that!!!