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Question Number 36911 by prof Abdo imad last updated on 07/Jun/18

$$pisapolynomehavingnrootssimples$$$$x_i(1⩽x_i⩽n)with{x}_i^2\ne 1calculste$$$$\sum _{k=1}^n\frac{1}{1-x_k}.$$

Commented by math khazana by abdo last updated on 09/Jun/18

$$wehavep\left(x\right)=\lambda \prod _{i=1}^n(x-x_i)and$$$$p^{{}^{\prime }}\left(x\right)=\lambda \sum _{i=1}^n\prod _{k=1andk\ne i}^n(x-x_k)\Rightarrow$$$$\frac{p^{{}^{\prime }}\left(x\right)}{p\left(x\right)}=\sum _{k=1}^n\frac{1}{x-x_k}\Rightarrow \sum _{k=1}^n\frac{1}{1-x_k}=\frac{p^{{}^{\prime }}\left(1\right)}{p\left(1\right)}.$$

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