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14 tháng 9 2015

2x+2x+1+2x+2+.....+2x+2020 = 22021 - 1

2x.(1+2+22+....+22020) = 2021 - 1

Đặt M = 1+2+22+...+22020

2M = 2+22+23+...+22021

2M - M = 22021-1

=> M = 22021 - 1

Thay vào, ta có:

2x.(22021 - 1) = 22021 - 1

=> 2x = 1

=> x = 0

x−42021+x−32020=x−22019+x−12018

⇔ x−42021+x−32020−x−22019−x−12018=0

⇔ (1+x−42021)+(1+x−32020)−(1+x−22019)−(1+x−12018)=0⇔ x+20172021+x+20172020−x+20172019−x+20172018=0

⇔ (x+2017)(12021+12020−12019−12018)=0

⇔ x + 2017 = 0

⇔ x = -2017

17 tháng 3 2020

\(\frac{x-1}{2020}+\frac{x-2}{2021}=\frac{x+1}{2018}+\frac{x+2}{2017}\)

\(\Leftrightarrow\frac{x-1}{2020}+1+\frac{x-2}{2021}-1=\frac{x+1}{2018}+1+\frac{x+2}{2017}+1\)

\(\Leftrightarrow\frac{x+2019}{2020}+\frac{x+2019}{2021}=\frac{x+2019}{2018}+\frac{x+2019}{2017}\)

\(\Leftrightarrow\left(x+2019\right)\left(\frac{1}{2020}+\frac{1}{2021}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)

mà \(\frac{1}{2020}+\frac{1}{2021}-\frac{1}{2018}-\frac{1}{2017}\ne0\)

\(\Leftrightarrow x+2019=0\)

\(\Leftrightarrow x=-2019\)

25 tháng 2 2023

\(\dfrac{x-4}{2022}+\dfrac{x-3}{2021}+\dfrac{x-2}{2020}+\dfrac{x-1}{2019}\text{=}-4\)

\(\dfrac{x-4}{2022}+\dfrac{x-3}{2021}+\dfrac{x-2}{2020}+\dfrac{x-1}{2019}+4\text{=}0\)

\(\left(\dfrac{x-4}{2022}+1\right)+\left(\dfrac{x-3}{2021}+1\right)+\left(\dfrac{x-2}{2020}+1\right)+\left(\dfrac{x-1}{2019}+1\right)\text{=}0\)

\(\dfrac{x-2018}{2022}+\dfrac{x-2018}{2021}+\dfrac{x-2018}{2020}+\dfrac{x-2018}{2019}\text{=}0\)

\(\left(x-2018\right)\left(\dfrac{1}{2022}+\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}\right)\text{=}0\)

\(Do:\) \(\dfrac{1}{2022}+\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}\ne0\)

\(x-2018\text{=}0\)

\(x\text{=}2018\)

\(Vậy...\)

\(x_1+x_2=x_3+x_4=...=x_{2019}+x_{2020}=2\Rightarrow x_1+x_2+x_3+x_4+...+x_{2019}+x_{2020}=2.1010=2020\)

\(\Rightarrow x_1+x_2+x_3+x_4+...+x_{2019}+x_{2020}+x_{2021}=2020+x_{2021}\)

\(\Rightarrow0=2020+x_{2021}\)

\(\Rightarrow x_{2021}=-2020\)

                                     Vậy \(x_{2021}=-2020\)

16 tháng 9 2020

Sửa đề: \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{2020}{2021}\) \(Đkxđ:\hept{\begin{cases}x\ne0\\x\ne-2\end{cases}}\)

\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2020}{2021}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{2020}{2021}\)

\(\Leftrightarrow\frac{x+2}{2021}=1\)

\(\Leftrightarrow x=2019\)

Vậy \(x=2019\)