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8 tháng 4 2022
refer
https://lazi.vn/edu/exercise/634984/tim-x-biet-x-1-2019-x-2-2020-x-3-2021x-4-2022
ND
2
21 tháng 8 2019
\(x-2019+\frac{x-2020}{2}=\frac{x-2021}{3}+\frac{x-2022}{4}\)
\(\Rightarrow x-2019+1+\frac{x-2020}{2}+1=\frac{x-2021}{3}+1+\frac{x-2022}{4}+1\)
\(\Rightarrow x-2018+\frac{x-2020+2}{2}=\frac{x-2021+3}{3}+\frac{x-2022+4}{4}\)
\(\Rightarrow x-2018+\frac{x-2018}{2}-\frac{x-2018}{3}-\frac{x-2018}{4}=0\)
\(\Rightarrow\left(x-2018\right)\left(1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)=0\)
\(\Rightarrow-\frac{1}{12}\left(x-2018\right)=0\Leftrightarrow x=2018\)
B
4 tháng 9 2020
Bài làm :
Ta có :
\(x-2019+\frac{x-2020}{2}=\frac{x-2021}{3}+\frac{x-2022}{4}\)
\(\Rightarrow x-2019+1+\frac{x-2020}{2}+1=\frac{x-2021}{3}+1+\frac{x-2022}{4}+1\)
\(\Rightarrow x-2018+\frac{x-2020+2}{2}=\frac{x-2021+3}{3}+\frac{x-2022+4}{4}\)
\(\Rightarrow x-2018+\frac{x-2018}{2}-\frac{x-2018}{3}-\frac{x-2018}{4}=0\)
\(\Rightarrow\left(x-2018\right)\left(1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)=0\)
\(\text{Vì : }\left(1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)\ne0\Rightarrow x-2018=0\)
\(\Rightarrow x=2018\)
Vậy x=2018
NH
0
15 tháng 6 2023
B/A
\(=\dfrac{1+\dfrac{2020}{2}+1+\dfrac{2019}{3}+...+1+\dfrac{1}{2021}+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}\)
\(=\dfrac{2022\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}=2022\)
HK
16 tháng 4 2023
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = (\(\dfrac{2021}{2}+1\))+(\(\dfrac{2020}{3}+1\))+....+(\(\dfrac{1}{2022}+1\))
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = \(\dfrac{2023}{2}\)+\(\dfrac{2023}{3}\)+....+ \(\dfrac{2023}{2022}\)
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = 2023.( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\))
vậy x= 2023
PN
0
5 tháng 2 2020
\(x_1+x_2=x_3+x_4=...=x_{2019}+x_{2020}=2\Rightarrow x_1+x_2+x_3+x_4+...+x_{2019}+x_{2020}=2.1010=2020\)
\(\Rightarrow x_1+x_2+x_3+x_4+...+x_{2019}+x_{2020}+x_{2021}=2020+x_{2021}\)
\(\Rightarrow0=2020+x_{2021}\)
\(\Rightarrow x_{2021}=-2020\)
Vậy \(x_{2021}=-2020\)
NH
3
H
19 tháng 5 2021
1. \(\dfrac{2019}{2020}-\left(\dfrac{2019}{2020}-\dfrac{2020}{2021}\right)\)
\(=\dfrac{2019}{2020}-\dfrac{2019}{2020}+\dfrac{2020}{2021}\)
\(=0+\dfrac{2020}{2021}=\dfrac{2020}{2021}\)
19 tháng 5 2021
Giải:
1) \(\dfrac{2019}{2020}-\left(\dfrac{2019}{2020}-\dfrac{2020}{2021}\right)\)
\(=\dfrac{2019}{2020}-\dfrac{2019}{2020}+\dfrac{2020}{2021}\)
\(=\left(\dfrac{2019}{2020}-\dfrac{2019}{2020}\right)+\dfrac{2020}{2021}\)
\(=0+\dfrac{2020}{2021}\)
\(=\dfrac{2020}{2021}\)
2) \(\dfrac{2}{9}+\dfrac{7}{9}:\left(\dfrac{42}{5}-\dfrac{7}{5}\right)\)
\(=\dfrac{2}{9}+\dfrac{7}{9}:7\)
\(=\dfrac{2}{9}+\dfrac{1}{9}\)
\(=\dfrac{1}{3}\)
3) \(\dfrac{3}{4}+\dfrac{x}{4}=\dfrac{5}{8}\)
\(\dfrac{x}{4}=\dfrac{5}{8}-\dfrac{3}{4}\)
\(\dfrac{x}{4}=\dfrac{-1}{8}\)
\(\Rightarrow x=\dfrac{4.-1}{8}=\dfrac{-1}{2}\)
4) \(\left|3x+1\right|-\dfrac{1}{4}=\dfrac{-1}{4}\)
\(\left|3x-1\right|=\dfrac{-1}{4}+\dfrac{1}{4}\)
\(\left|3x-1\right|=0\)
\(3x-1=0\)
\(3x=0+1\)
\(3x=1\)
\(x=1:3\)
\(x=\dfrac{1}{3}\)
Chúc bạn học tốt!
7 tháng 5 2023
=1+(2-3-4+5)+(6-7-8+9)+.....+(2018-2019-2020+2021)+2022
=1+0+0+.....+0+2022
=2023
số năm nay luôn
\(\dfrac{x-4}{2022}+\dfrac{x-3}{2021}+\dfrac{x-2}{2020}+\dfrac{x-1}{2019}\text{=}-4\)
\(\dfrac{x-4}{2022}+\dfrac{x-3}{2021}+\dfrac{x-2}{2020}+\dfrac{x-1}{2019}+4\text{=}0\)
\(\left(\dfrac{x-4}{2022}+1\right)+\left(\dfrac{x-3}{2021}+1\right)+\left(\dfrac{x-2}{2020}+1\right)+\left(\dfrac{x-1}{2019}+1\right)\text{=}0\)
\(\dfrac{x-2018}{2022}+\dfrac{x-2018}{2021}+\dfrac{x-2018}{2020}+\dfrac{x-2018}{2019}\text{=}0\)
\(\left(x-2018\right)\left(\dfrac{1}{2022}+\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}\right)\text{=}0\)
\(Do:\) \(\dfrac{1}{2022}+\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}\ne0\)
\(x-2018\text{=}0\)
\(x\text{=}2018\)
\(Vậy...\)