Răspuns:
Aducem la numitor comun, prin amplificare
[tex]E(x) = \bigg[\bigg(^{x^2-1)}1 - \dfrac{3x^2}{1 - x^2}\bigg) \cdot \dfrac{x - 1}{2x + 1} - 1\bigg] \cdot \dfrac{x^2 + 2x + 1}{x - 1}\\[/tex]
- 1 - x² = -(x² - 1) = -(x - 1)(x + 1)
- x² + 2x + 1 = (x + 1)²
Expresia devine
[tex]E(x) = \bigg[\dfrac{x^2 - 1 + 3x^2}{(x - 1)(x + 1)} \cdot \dfrac{x - 1}{2x + 1} - 1\bigg] \cdot \dfrac{(x + 1)^2}{x - 1}\\[/tex]
[tex]E(x) = \bigg[\dfrac{4x^2 - 1}{x + 1} \cdot \dfrac{1}{2x + 1} - 1 \bigg] \cdot \dfrac{(x + 1)^2}{x - 1}\\[/tex]
- 4x² - 1 = (2x)² - 1² (2x - 1)(2x + 1)
[tex]E(x) = \bigg[\dfrac{(2x - 1)(2x + 1)}{x + 1} \cdot \dfrac{1}{2x + 1} - \dfrac{x + 1}{x + 1} \bigg] \cdot \dfrac{(x + 1)^2}{x - 1}\\[/tex]
- simplificăm prin (2x + 1)
[tex]E(x) = \dfrac{2x - 1 - (x + 1)}{x + 1} \cdot \dfrac{(x + 1)^2}{x - 1} = \dfrac{x - 2}{x + 1} \cdot \dfrac{(x + 1)^2}{x - 1}\\[/tex]
[tex]E(x) = \dfrac{(x - 2)(x + 1)}{x - 1}\\[/tex]
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Pentru:
[tex]E(x) = \dfrac{1}{x + 1}[/tex]
[tex]E(3) = \dfrac{1}{3 + 1} = \dfrac{1}{4}; \ E(4) = \dfrac{1}{4 + 1} = \dfrac{1}{5}; \ ... ; \ E(9) = \dfrac{1}{9 + 1} = \dfrac{1}{10}\\[/tex]
[tex]\dfrac{1}{E(3)} + \dfrac{1}{E(4)} + ... + \dfrac{1}{E(9)} = 4 + 5 + ... + 10 = 49 > \dfrac{1}{2}[/tex]
