Question

Ex ăsta vă rog….. vreau rezolvare completă!

Answer 1

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Answer 2

Răspuns:

Fie $f(x)=x^n-3x^{n-1}+2x+1$ și $x_1, \ x_2, \ \ldots, x_n$ rădăcinile sale. Atunci

$f(x)=\left(x-x_1\right)\left(x-x_2\right)\ldots\left(x-x_n\right)$

$f'(x)=(x-x_2)(x-x_3)\ldots(x-x_n)+(x-x_1)(x-x_3)\ldots(x-x_n)+\ldots+(x-x_1)(x-x_2)\ldots(x-x_{n-1})$

Atunci

$\displaystyle\frac{f'(x)}{f(x)}=\displaystyle\frac{1}{x-x_1}+\frac{1}{x-x_2}+\ldots\frac{1}{x-x_n}$

Avem

$S=\displaystyle\sum_{k=1}^n\frac{x_k}{x_k-1}=\sum_{k=1}^n\frac{x_k-1+1}{x_k-1}=\sum_{k=1}^n\left(1+\frac{1}{x_k-1}\right)=n-\sum_{k=1}^n\frac{1}{1-x_k}}$

Dar

$\displaystyle\sum_{k=1}^n\frac{1}{1-x_k}=\frac{f'(1)}{f(1)}=-2n+5$

Rezultă $S=3n-5$.

Explicație pas cu pas: