Question

dau coroana


se considera E (x) = (2×-1)² -3(x-3)(x+2)-(x-2)(x+1)


determinați ca

E (1)1+E(1 ) +E 1 +.........+ E( 1)
2 3 2010 =4242​

Answer 1

Răspuns:

E(x) =4x^2-4x+1-3(x^2-x-6) - x^2+x+2

E(x) =4x^2-4x+1-3x^2+3x+18-x^2+x+2

E(x) =21

21x2020=42420

Answer 2

$\it E(x)=(2x-1)^2-3(x-3)(x+2)-(x-2)(x+1)=\\ \\ =4x^2-4x+1-3(x^2+2x-3x-6)-(x^2+x-2x-2)=\\ \\ =4(x^2-x)+1-3(x^2-x-6)-(x^2-x-2)$

$\it Not\breve am\ \ x^2-x=t, iar\ expresia\ devine:\\ \\ E=4t+1-3t+18-t+2=21\\ \\ Deci, E(x)=21,\ \ \forall\ x\in\mathbb{R}$

$\it E(1) +E(\dfrac{1}{2})+E(\dfrac{1}{3})+\ ...\ +E(\dfrac{1}{2020})=\underbrace{21+21+21+\ ...\ +21}_{2020\ termeni}=\\ \\ \\ =2020\cdot21=42420$