Question

va rog mult !!!!! :-)

Answer 1

a) E(3;1)= (2*3-1)^2-3*3+1=25-9+1=17
b) aici faci la fel ca la punctul a) ; inlocuiesti x,y cu valorile date in paranteze si calculezi
c) E(x;0)=0
4x^2 -3x+1=0
ec de grd 2 cu 2 radacini
o rezolvi cu delta

Answer 2

Salut,

E(x;y) = (2x-y)²-3x+1.

a) E(3;1) = (2•3-1)²-3•3+1
E(3;1) = (6-1)²-9+1
E(3;1) = 5²-8
E(3;1) = 25-8
E(3;1) = 17.

b) E(-1;1) = [2•(-1)-1]²-3•(-1)+1
E(-1;1) = (-2-1)²+3+1
E(-1;1) = (-3)²+4
E(-1;1) = 9+4
E(-1;1) = 13.

E(1;-1) = [2•1-(-1)]²-3•1+1
E(1;-1) = (2+1)²-3+1
E(1;-1) = 3²-2
E(1;-1) = 6-2
E(1;-1) = 4.

E(-2;1) = [2•(-2)-1]²-3•(-2)+1
E(-2;1) = (-4-1)²+6+1
E(-2;1) = (-5)²+7
E(-2;1) = 25+7
E(-2;1) = 32.

E(2;-1) = [2•2-(-1)]²-3•2+1
E(2;-1) = (4+1)²-6+1
E(2;-1) = 5²-5
E(2;-1) = 25-5
E(2;-1) = 20.


E(-1;1)+E(1;-1)+E(-2;1)+E(2;-1) = 13+4+32+20 = 17+52 = 69.


c) E(x;0) = 0
(2x-0)²-3x+1 = 0
4x²-3x+1 = 0.

a = 4
b = -3
c = 1

∆ = b²-4ac
∆ = 9-16
∆ = -7

∆ nu are rădăcini reale, deci S = ∅.


Succes! :3