a) x + 1 = 0 ⇒ x = - 1
x³ + 1 = 0 ⇒ x³ = -1 ⇒ x = -1
x² - x + 1 = 0, Δ = 1 - 4 = - 3 < 0 ⇒ x ∉ R
⇒ E(x) nu are valoare definită pentru x = - 1
b) Avem x³ + 1 = (x + 1)(x² -x + 1). Amplificăm fracțiile:
[tex]E(x) = \dfrac{x^{2} - x + 1}{(x + 1)(x^{2} - x + 1)} - \dfrac{2 + 2x}{x^{3} + 1} + \dfrac{2(x + 1)}{(x + 1)(x^{2} - x + 1)}\\[/tex]
[tex]= \dfrac{x^{2} - x + 1 - (2 + 2x) + 2(x + 1)}{(x + 1)(x^{2} - x + 1)} = \dfrac{x^{2} - x + 1 - 2 - 2x + 2x + 2}{(x + 1)(x^{2} - x + 1)}\\[/tex]
[tex]= \dfrac{x^{2} - x + 1}{(x + 1)(x^{2} - x + 1)} = \dfrac{1}{x + 1}[/tex]
Arunci:
[tex](x + 1) \cdot E(x) = (x + 1) \cdot \dfrac{1}{x + 1} = \bf 1[/tex]
[tex]c) \ 3 \cdot E(x) = \dfrac{1}{7} \implies \dfrac{3}{x + 1} = \dfrac{1}{7} \implies x + 1 = 21 \implies \bf x = 20[/tex]
[tex]d) \ 6 \cdot E(x) \in \Bbb{Z} \implies \dfrac{6}{x + 1} \in \Bbb{Z} \implies (x + 1) \in \{-6,-3,-2,-1,1,2,3,6\}[/tex]
[tex]\implies x \in \{-5,-4,-3,-2,0,1,2,5\}[/tex]
[tex]\boxed{\boldsymbol{\dfrac{1}{k \cdot (k + 1)} = \dfrac{1}{k} - \dfrac{1}{k + 1}}}[/tex]
[tex]e) \ \dfrac{1}{2} \cdot \dfrac{1}{3} + \dfrac{1}{3} \cdot \dfrac{1}{4} + ... + \dfrac{1}{20} \cdot \dfrac{1}{21} = \\[/tex]
- se aplică formula de recurență:
[tex]= \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{20} - \dfrac{1}{21}\\[/tex]
- se reduc termenii asemenea
[tex]\dfrac{1}{2} - \dfrac{1}{21} = \dfrac{21 - 2}{2 \cdot 21} = \bf \dfrac{19}{42}[/tex]
