Question

Salutare !
Eu nu pot sa rezolv corect acest exercitiu(atasez pdf), si daca aveti, dumneavoastra amabilitatea,as aprecia.

20e5399d24ff6f7ba6128e79b726b8ab.pdf

Answer 1

$(-a)^{2k} = a^{2k},\quad \forall k\in\mathbb{Z}$

$f:\mathbb{R}\to \mathbb{R},\quad f(x) = x^{2012}+x^2+1 \\ \\\\ f(-3)-f(-1)+f(1)-f(3) = \\ \\ = \Big[(-3)^{2012}+(-3)^2+1\Big] - \Big[(-1)^{2012}+(-1)^2+1\Big]+1^{2012}+1^2+1 - \\ -(3^{2012}+3^2+1) = \\ \\ = (3^{2012}+3^2+1) - 3+3-(3^{2012}+3^2+1)=\\ \\=(3^{2012}+3^2+1)-(3^{2012}+3^2+1)=\\ \\ = 0$

Answer 2

f : R → R

f(x) = x²⁰¹² + x² + 1

f(-3) - f(-1) + f(1) - f(3) = 0

f(-3) = (-3)²⁰¹² + (-3)² + 1 = 3²⁰¹² + 10

f(-1) = (-1)²⁰¹² + (-1)² + 1 = 1 + 1 + 1 = 3

f(1) = 1²⁰¹² + 1² + 1 = 1 + 1 + 1 = 3

f(3) = 3²⁰¹² + 3² + 1 = 3²⁰¹² + 10

=> f(-3) - f(-1) + f(1) - f(3) = ( 3²⁰¹² + 10) - 3 + 3 - (3²⁰¹² + 10)=

= 3²⁰¹² + 10 + 0 - 3²⁰¹² - 10 = 0 ( se reduc termenii cu semn opus )