Tìm x:
x-1/2020+x-2/2021=x+1/2018+x+2/2017
K
Khách
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Những câu hỏi liên quan
KL
4 tháng 8 2017
a, \(\dfrac{2017.2021-4031}{2020+2017.2018}\)
= \(\dfrac{2017\left(2018+3\right)-4031}{2020+2017.2018}\)
= \(\dfrac{2017.2018+2017.3-4031}{2020+2017.2018}\)
= \(\dfrac{2017.2018+2020}{2020+2017.2018}\)
= 1
@Nguyen Thi Ngoc Linh
NN
3
17 tháng 9 2020
\(\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2017}=\frac{x-1}{2021}+\frac{x-2}{2022}+\frac{x-3}{2023}\)
\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+\left(\frac{x+3}{2017}+1\right)=\left(\frac{x-1}{2021}+1\right)+\left(\frac{x-2}{2022}+1\right)+\left(\frac{x-3}{2023}+1\right)\)
\(\Leftrightarrow\left(\frac{x+1+2019}{2019}\right)+\left(\frac{x+2+2018}{2018}\right)+\left(\frac{x+3+2017}{2017}\right)=\left(\frac{x-1+2021}{2021}\right)+\left(\frac{x-2+2022}{2022}\right)+\left(\frac{x-3+2023}{2023}\right)\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2021}+\frac{x+2020}{2022}+\frac{x+2020}{2023}\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}-\frac{x+2020}{2021}-\frac{x+2020}{2022}-\frac{x+2020}{2023}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\right)=0\)
Vì \(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\ne0\)
=> x + 2020 = 0
=> x = -2020
B
17 tháng 9 2020
Bài làm :
Ta có :
\(\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2017}=\frac{x-1}{2021}+\frac{x-2}{2022}+\frac{x-3}{2023}\)
\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+\left(\frac{x+3}{2017}+1\right)=\left(\frac{x-1}{2021}+1\right)+\left(\frac{x-2}{2022}+1\right)+\left(\frac{x-3}{2023}+1\right)\)
\(\Leftrightarrow\left(\frac{x+1+2019}{2019}\right)+\left(\frac{x+2+2018}{2018}\right)+\left(\frac{x+3+2017}{2017}\right)=\left(\frac{x-1+2021}{2021}\right)+\left(\frac{x-2+2022}{2022}\right)+\left(\frac{x-3+2023}{2023}\right)\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2021}+\frac{x+2020}{2022}+\frac{x+2020}{2023}\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}-\frac{x+2020}{2021}-\frac{x+2020}{2022}-\frac{x+2020}{2023}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\right)=0\)
\(\text{Vì : }\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\ne0\)
\(\Rightarrow x+2020=0\Leftrightarrow x=-2020\)
Vậy x=-2020
KL
1
8 tháng 8 2017
\(\dfrac{x+2017}{x+2018}=\dfrac{2020}{2021}\)
\(\Leftrightarrow1-\dfrac{x+2017}{x+2018}=1-\dfrac{2020}{2021}\)
\(\Leftrightarrow\dfrac{x+2018}{x+2018}-\dfrac{x+2017}{x+2018}=\dfrac{2021}{2021}-\dfrac{2020}{2021}\)
\(\Leftrightarrow\dfrac{\left(x+2018\right)-\left(x+2017\right)}{x+2018}=\dfrac{2021-2020}{2021}\)
\(\Leftrightarrow\dfrac{x+2018-x-2017}{x+2018}=\dfrac{1}{2021}\)
\(\Leftrightarrow\dfrac{\left(2018-2017\right)+\left(x+x\right)}{x+2018}=\dfrac{1}{2021}\)
\(\Leftrightarrow\dfrac{1}{x+2018}=\dfrac{1}{2021}\)
\(\Leftrightarrow x+2018=2021\)
\(\Leftrightarrow x=3\left(tm\right)\)
vậy ....
LC
9
22 tháng 3 2023
\(\dfrac{x-1}{2021}+\dfrac{x-2}{2020}=\dfrac{x-5}{2017}+\dfrac{x-7}{2015}\\ \dfrac{x-1}{2021}+\dfrac{x-2}{2020}-2=\dfrac{x-5}{2017}+\dfrac{x-7}{2015}-2\\ \dfrac{x-1}{2021}+\dfrac{x-2}{2020}-1-1=\dfrac{x-5}{2017}+\dfrac{x-7}{2015}-1-1\\\left(\dfrac{x-1}{2021}-1\right)+\left(\dfrac{x-2}{2020}-1\right)=\left(\dfrac{x-5}{2017}-1\right)+\left(\dfrac{x-7}{2015}-1\right)\\ \dfrac{x-2022}{2021}+\dfrac{x-2022}{2020}=\dfrac{x-2022}{2017}+\dfrac{x-2022}{2015}\\ \dfrac{x-2022}{2021}+\dfrac{x-2022}{2020}-\dfrac{x-2022}{2017}-\dfrac{x-2022}{2015}=0\\ \left(x-2022\right)\left(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2017}-\dfrac{1}{2015}\right)=0\)
mà `(1/2021+1/2020-1/2017-1/2015 \ne 0`
nên `x-2022=0`
`x=2022`
XO
1 tháng 10 2020
Ta có :\(\frac{x+4}{2018}+\frac{x+3}{2019}=\frac{x+2}{2020}+\frac{x+1}{2021}\)
=> \(\left(\frac{x+4}{2018}+1\right)+\left(\frac{x+3}{2019}+1\right)=\left(\frac{x+2}{2020}+1\right)+\left(\frac{x+1}{2021}+1\right)\)
=> \(\frac{x+2022}{2018}+\frac{x+2022}{2019}=\frac{x+2022}{2020}+\frac{x+2022}{2021}\)
=> \(\frac{x+2022}{2018}+\frac{x+2022}{2019}-\frac{x+2022}{2020}-\frac{x+2022}{2021}=0\)
=> \(\left(x+2022\right)\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\right)=0\)
Vì \(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\ne0\)
=> x + 2022 = 0
=> x = -2022
Vậy x = -2022
1 tháng 10 2020
\(\frac{x+4}{2018}+\frac{x+3}{2019}=\frac{x+2}{2020}+\frac{x+1}{2021}\)
\(\frac{x+4}{2018}+1+\frac{x+3}{2019}+1=\frac{x+2}{2020}+1+\frac{x+1}{2021}+1\)
\(\frac{x+4}{2018}+\frac{2018}{2018}+\frac{x+3}{2019}+\frac{2019}{2019}=\frac{x+2}{2020}+\frac{2020}{2020}+\frac{x+1}{2021}+\frac{2021}{2021}\)
\(\frac{x+2022}{2018}+\frac{x+2022}{2019}=\frac{x+2022}{2020}+\frac{x+2022}{2021}\)
\(\frac{x+2022}{2018}+\frac{x+2022}{2019}-\frac{x+2022}{2020}-\frac{x+2022}{2021}=0\)
\(\left(x+2022\right)\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\right)=0\)
\(x+2022=0\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\ne0\right)\)
\(x=0-2022\)
\(x=-2022\)
15 tháng 9 2019
\(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}+\frac{x+4}{2017}=-4\)
=> \(\left[\frac{x+1}{2020}+1\right]+\left[\frac{x+2}{2019}+1\right]+\left[\frac{x+3}{2018}+1\right]+\left[\frac{x+4}{2017}+1\right]=-4\)
=> \(\left[\frac{x+1}{2020}+\frac{2020}{2020}\right]+\left[\frac{x+2}{2019}+\frac{2019}{2019}\right]+\left[\frac{x+3}{2018}+\frac{2018}{2018}\right]+\left[\frac{x+4}{2017}+\frac{2017}{2017}\right]=-4\)
=> \(\frac{x+2021}{2020}+\frac{x+2021}{2019}+\frac{x+2021}{2018}+\frac{x+2021}{2017}=-4\)
=> \(\left[x+2021\right]\left[\frac{1}{2000}+\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}\right]=-4\)
Do \(\frac{1}{2020}>\frac{1}{2019}>\frac{1}{2018}>\frac{1}{2017}\)nên \(\frac{1}{2000}+\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}\ne0\)
Do đó : x + 2021 = -4 => x = -4 - 2021 = -2025
x−42021+x−32020=x−22019+x−12018x−42021+x−32020=x−22019+x−12018
⇔ x−42021+x−32020−x−22019−x−12018=0x−42021+x−32020−x−22019−x−12018=0
⇔ (1+x−42021)+(1+x−32020)−(1+x−22019)−(1+x−12018)=0(1+x−42021)+(1+x−32020)−(1+x−22019)−(1+x−12018)=0⇔ x+20172021+x+20172020−x+20172019−x+20172018=0x+20172021+x+20172020−x+20172019−x+20172018=0
⇔ (x+2017)(12021+12020−12019−12018)=0(x+2017)(12021+12020−12019−12018)=0
⇔ x + 2017 = 0
⇔ x = -2017
\(\frac{x-1}{2020}+\frac{x-2}{2021}=\frac{x+1}{2018}+\frac{x+2}{2017}\)
\(\Leftrightarrow\frac{x-1}{2020}+1+\frac{x-2}{2021}-1=\frac{x+1}{2018}+1+\frac{x+2}{2017}+1\)
\(\Leftrightarrow\frac{x+2019}{2020}+\frac{x+2019}{2021}=\frac{x+2019}{2018}+\frac{x+2019}{2017}\)
\(\Leftrightarrow\left(x+2019\right)\left(\frac{1}{2020}+\frac{1}{2021}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)
mà \(\frac{1}{2020}+\frac{1}{2021}-\frac{1}{2018}-\frac{1}{2017}\ne0\)
\(\Leftrightarrow x+2019=0\)
\(\Leftrightarrow x=-2019\)