xy-x-y+2,=(x-1)(y-1)+1
x⊥x=(x-1)(y-1)+1
(x-1)(x-1)+1≤2
(x-1)(x-1)≤1
x²-1≤1
x²≤2
x∈[-√2;√2]
3.
a) calcul direct
2( x-2) (y-2) +2 =2(xy-2x-2y+4)+2=2xy-4x-4y+8+2=2xy-4x-4y+10, adevarat
b)x⊥e=2( x-2) (e-2) +2=x
2( x-2) (e-2) +2-x=0
2( x-2) (e-2) -(x-2)=0
(x-2)[2(e-2)-1]=0
(x-2)(2e-5)=0 ∀x ∈ R
⇒2e-5=0
e=5/2
3c)
( -10) o (-9) o....o 0 o 1 o...o 9 o 10. = (-10)°(-9)......°(1)°(2)°(3)°.....°(10)=
a°2°b in care cu a am notat (-10)°....°1
si cu b=2°3°....°10 conmsiderand legea ASOCIATIVAS 9ceea fie tebuyia spus in ipotrza, fie bagata ca cerinta in punctele anterioare; legea ESTE asocoiativa, dar nu mai demonstrez, ca sa nu lungesc rezolvarea)
legea este si comutativa , se poate arata usor
y°x= 2(y-2)(x-2)+2=x°y
se arata ca ∀x, x°2=2°x=2
intr-adevar
x°2= 2(x-2)(2-2)+2=0+2=2
deci a°2°b=2°b=2
deci
( -10) o (-9) o....o 0 o 1 o...o 9 o 10. =2
