tex]...

August 2022 1 58 Report

Функция f удовлетворяет равенству [tex](x-1)f(x)+f(\frac{1}{x} ) = \frac{1}{x-1}[/tex] для каждого значения x, не равного 0 и 1. Найдите [tex]f(\frac{2020}{2021} )[/tex]

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igorShap

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Ответ:

$f(\frac{2020}{2021})=2021$

Объяснение:

$x=\frac{2020}{2021}\Rightarrow -\frac{1}{2021}f(\frac{2020}{2021})+f(\frac{2021}{2020})=-2021\;\;\;\;\;\;\;\;\;(1)\\ x=\frac{2021}{2020}\Rightarrow \;\;\:\frac{1}{2020}f(\frac{2021}{2020})+f(\frac{2020}{2021})=\;\;\:2020\;\;\;\;\;\;\;\;\;(2)$

Найдем разность $(1)-2020\cdot (2)$ :

$-\frac{1}{2021}f(\frac{2020}{2021})+f(\frac{2021}{2020})-2020(\frac{1}{2020}f(\frac{2021}{2020})+f(\frac{2020}{2021}))=-2021-2020\cdot 2020\\ -\frac{1}{2021}f(\frac{2020}{2021})+f(\frac{2021}{2020})-f(\frac{2021}{2020})-2020 f(\frac{2020}{2021})=-2021-2020^2\\ -\frac{1}{2021}f(\frac{2020}{2021})-2020 f(\frac{2020}{2021})=-(2021\cdot 2020+1)$

$\frac{2021\cdot 2020+1}{2021}f(\frac{2020}{2021})=2021\cdot 2020+1\\ f(\frac{2020}{2021})=2021$

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