Question

exercitiul 5 va rog frumos dau coroana

Answer 1

E(x)=(x²-x+2)²-(x²-x)²-(2x-1)²

E(x)=(x(x-1)+2)²-(x(x-1))²-(2x-1)²

E(x)=(x²-x+2)²-x²(x-1)²-(2x-1)²

E(x)=x⁴-x³+2x²-x³+x²-2x+2x²-2x+4-x⁴+2x³-x²-(2x)²+2×2x-1

E(x)=x⁴-x³+2x²-x³+x²-2x+2x²-2x+4-x⁴+2x³-x²-4x²+4x-1

E(x)=(x⁴-x⁴)+(-x³-x³+2x³)+(2x²+x²+2x²-x²-4x²)+(-2x-2x+4x)+(4-1)

E(x)=3

Answer 2

Formule de calcul prescurtat aplicate in exercitiul dat:

(a+b+c)²=a²+b²+c²+2ab+2ac+2bc

(a-b)²=a²-2*a*b+b²

Explicație pas cu pas:

E(X)=(x²-x+2)²-(x²-x)²-(2x-1)²=(x²)²+(-x)²+(2)²+2*x²*(-x)+2*x²*2+2*(-x)*2-((x²)²-2*x²*x+x²)-((2x)²-2*2x*1+1²)=x⁴+x²+4-2x³+4x²-4x-(x⁴-2x³+x²)-(4x²-4x+1)=

x⁴+x²+4-2x³+4x²-4x-x⁴+2x³-x²-4x²+4x-1=3.

Deci am demonstrat ca E(X)=3 pentru orice x∈R.

Bafta!