Question

Mate, clasa a 8-a
ajutor și seriozitate rog
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Answer 1

$E(x) = \dfrac{1}{x^{2} + x} + \dfrac{1}{x^{2} + 3x + 1 \cdot 2} + \dfrac{1}{x^{2} + 5x + 2 \cdot 3} + ... + \dfrac{1}{x^{2} + 4021x + 2010 \cdot 2011}$

din Relațiile lui Viete:

$x^{2} -Sx+P=0 \iff (x - x_{1})(x - x_{2}) = 0$

​ $\begin{cases}S=x_{1}+x_{2} \\P=x_{1} \cdot x_{2} \end{cases}$

numitorii sunt:

$x^{2} + x = 0 \iff x(x+1) = 0$

$\iff x=-1;x=0$

$x^{2} + 3x +1 \cdot 2 = x^{2} + (1+2)x +1 \cdot 2 = (x+1)(x+2)$

$\iff (x+1)(x+2)=0 \iff x=-2;x=-1$

$x^{2} + 5x + 2 \cdot 3 = x^{2} + (2+3)x + 2 \cdot 3 = (x+2)(x+3)$$\iff (x+2)(x+3)=0 \iff x=-3;x=-2$

...

$x^{2} + 4021x + 2010 \cdot 2011 = x^{2} + (2010+2011)x + 2010 \cdot 2011 = (x+2010)(x+2011)$$\iff (x+2010)(x+2011)=0 \iff x=-2011;x=-2010$

a) expresia are sens pentru:

$x \in \mathbb{R} / \{-2011;-2010;...;-3;-2;-1;0\}$

b) cunoaștem formula:

$\dfrac{1}{x^{2} + x} = \dfrac{1}{x(x + 1)} = \dfrac{1}{x} - \dfrac{1}{x + 1}$

putem scrie expresia:

$E(x) = \bigg(\dfrac{1}{x} - \dfrac{1}{x+1}\bigg) + \bigg(\dfrac{1}{x+1} - \dfrac{1}{x+2}\bigg) + \bigg(\dfrac{1}{x+2} - \dfrac{1}{x+3}\bigg) +...+ \bigg(\dfrac{1}{x+2010} - \dfrac{1}{x+2011}\bigg) = \dfrac{1}{x} - \dfrac{1}{x+2011} = \dfrac{x+2011-x}{x(x+2011)} = \dfrac{2011}{x(x+2011)}$$\implies E(x) = \dfrac{2011}{x(x+2011)}$

c)

$\dfrac{2011}{E(x)} = x + 2011 \iff E(x)(x + 2011) = 2011$

$\dfrac{2011}{x(x+2011)} \cdot (x + 2011) = 2011$

$\dfrac{1}{x} = 1 \implies \bf x = 1$