Question

Se da polinomul P.
14 si 15, are cineva o idee?

Answer 1

Pe 14) l-a rezolvat Halogenhalogen.

15)

$\dfrac{1}{x_k-x_k^2}= -\dfrac{1}{x_k^2-x_k}= -\dfrac{1}{x_k(x_k-1)}= -\Big(\dfrac{1}{x_k-1}-\dfrac{1}{x_k}\Big) = \\ \\ = \dfrac{1}{x_k}-\dfrac{1}{x_k-1}\\ \\\\ \displaystyle \sum\limits_{k=1}^{20}\dfrac{1}{x_k-x_k^2} = \sum\limits_{k=1}^{20}\Big(\dfrac{1}{x_k}-\dfrac{1}{x_k-1}\Big) = \\ \\ = \dfrac{1}{x_1}+\dfrac{1}{x_2}+...+\dfrac{1}{x_{20}}-\Big(\dfrac{1}{x_1-1}+\dfrac{1}{x_2-1}+...+\dfrac{1}{x_{20}-1}\Big)=$

$\displaystyle = \dfrac{S_{19}}{S_{20}}+\sum\limits_{k=1}^{20} \dfrac{1}{1-x_k}= \\ \\ = \dfrac{\pm\dfrac{0}{1}}{\pm\dfrac{2}{1}}+\sum\limits_{k=1}^{20} \dfrac{1}{1-x_k} = \sum\limits_{k=1}^{20} \dfrac{1}{1-x_k}\\ \\\\\\f(x) = (x-x_1)(x-x_2)\cdot...\cdot (x-x_{20})$

$\displaystyle f'(x) = \sum\limits_{k=1}^{20}\dfrac{f(x)}{x-x_k}\\ \\ f'(1) = \sum\limits_{k=1}^{20}\dfrac{f(1)}{1-x_k}= \sum\limits_{k=1}^{20}\dfrac{5}{1-x_k}\\ \\ 20\cdot 1^{19}+10\cdot 1^{9}+5\cdot 1^4 = \sum\limits_{k=1}^{20}\dfrac{5}{1-x_k}\\ \\ 35 = 5\sum\limits_{k=1}^{20}\dfrac{1}{1-x_k}$

$\displaystyle\Rightarrow \sum\limits_{k=1}^{20}\dfrac{1}{1-x_k} = \dfrac{35}{5} = 7\\ \\ \\ \Rightarrow\boxed{\sum\limits_{k=1}^{20}\dfrac{1}{x_k-x_k^2} = 7}$

Answer 2

Răspuns:

14

Explicație pas cu pas: