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BL
0
BL
1
O
0
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
\(\frac{x+4}{2018}+\frac{x+3}{2019}=\frac{x+2}{2020}+\frac{x+1}{2021}\)
\(\frac{x+4}{2018}+1+\frac{x+3}{2019}+1=\frac{x+2}{2020}+1+\frac{x+1}{2021}+1\)
\(\frac{x+4}{2018}+\frac{2018}{2018}+\frac{x+3}{2019}+\frac{2019}{2019}=\frac{x+2}{2020}+\frac{2020}{2020}+\frac{x+1}{2021}+\frac{2021}{2021}\)
\(\frac{x+2022}{2018}+\frac{x+2022}{2019}=\frac{x+2022}{2020}+\frac{x+2022}{2021}\)
\(\frac{x+2022}{2018}+\frac{x+2022}{2019}-\frac{x+2022}{2020}-\frac{x+2022}{2021}=0\)
\(\left(x+2022\right)\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\right)=0\)
\(x+2022=0\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\ne0\right)\)
\(x=0-2022\)
\(x=-2022\)
CN
1 tháng 1 2018
\(\dfrac{x-4}{2021}+\dfrac{x-3}{2020}=\dfrac{x-2}{2019}+\dfrac{x-1}{2018}\)
⇔ \(\dfrac{x-4}{2021}+\dfrac{x-3}{2020}-\dfrac{x-2}{2019}-\dfrac{x-1}{2018}=0\)
⇔ \(\left(1+\dfrac{x-4}{2021}\right)+\left(1+\dfrac{x-3}{2020}\right)-\left(1+\dfrac{x-2}{2019}\right)-\left(1+\dfrac{x-1}{2018}\right)=0\)⇔ \(\dfrac{x+2017}{2021}+\dfrac{x+2017}{2020}-\dfrac{x+2017}{2019}-\dfrac{x+2017}{2018}=0\)
⇔ \(\left(x+2017\right)\left(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2019}-\dfrac{1}{2018}\right)=0\)
⇔ x + 2017 = 0
⇔ x = -2017
Vậy x = -2017
LC
9
NH
22 tháng 2 2023
a)
`(2x-1)(x+2/3)=0`
\(< =>\left[{}\begin{matrix}2x-1=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b)
\(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)
\(< =>\dfrac{x+4}{2019}+1+\dfrac{x+3}{2020}+1=\dfrac{x+2}{2021}+1+\dfrac{x+1}{2022}+1\)
\(< =>\dfrac{x+2023}{2019}+\dfrac{x+2023}{2020}=\dfrac{x+2023}{2021}+\dfrac{x+2023}{2022}\)
\(< =>\left(x+2023\right)\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\right)=0\)
\(< =>x+2023=0\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\ne0\right)\\ < =>x=-2023\)
BC
5
NH
16 tháng 7 2019
\(\frac{x+4}{2019}+\frac{x+3}{2020}=\frac{x+2}{2021}+\frac{x+1}{2020}\)
\(\Leftrightarrow(\frac{x+4}{2019}+1)+(\frac{x+3}{2020}+1)=(\frac{x+2}{2021}+1)+(\frac{x+1}{2022}+1)\)
\(\Leftrightarrow\frac{x+2023}{2019}+\frac{x+2023}{2020}=\frac{x+2023}{2021}+\frac{x+2023}{2022}\)
\(\Leftrightarrow\frac{x+2023}{2019}+\frac{x+2023}{2020}-\frac{x+2023}{2021}-\frac{x+2023}{2022}=0\)
\(\Leftrightarrow\left(x+2023\right)\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}-\frac{1}{2020}\right)=0\)
\(\Leftrightarrow x+2023=0\)
\(\Leftrightarrow x=-2023\)
Ta có: \(\frac{x+1}{2019}+\frac{x+1}{2020}=\frac{x+1}{2021}\)
\(\Leftrightarrow\frac{x+1}{2019}+\frac{x+1}{2020}-\frac{x+1}{2021}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}\right)=0\)
Mà \(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}>0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
\(\frac{x+1}{2019}+\frac{x+1}{2020}=\frac{x+1}{2021}\)
\(\frac{x+1}{2019}+\frac{x+1}{2020}-\frac{x+1}{2021}=0\)
\(\left(x+1\right)\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}\right)=0\)
\(x+1=0\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}\ne0\right)\)
\(x=0-1=-1\)