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LA
1
Hình như là vậy á
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
K
1
CT
0
KK
2
KK
14 tháng 2 2019
x2019-2019.x2018+2019.x2018+2019.x2017-2019.x2016+......2019.x-200 Tại x=2018
Giúp mik vs nhé
14 tháng 2 2019
Sai đề nên t sửa luôn nhé!
Vì \(x=2018\Rightarrow2019=2018+1=x+1\)
\(A=x^{2017}-2019\cdot x^{2018}+2019\cdot x^{2017}-2019\cdot x^{2016}+....+2019\cdot x-200\)
\(\Rightarrow A=x^{2019}-\left(x+1\right)x^{2018}+\left(x+1\right)x^{2017}-\left(x+1\right)x^{2016}+....-\left(x+1\right)x^2+\left(x+1\right)x-200\)
\(\Rightarrow A=x^{2019}-x^{2019}-x^{2018}+x^{2018}+x^{2017}-x^{2017}-x^{2016}+....-x^3-x^2+x^2+x-200\)
\(\Rightarrow A=x-200=2018-200=1818\)
PH
0
LV
0
\(\dfrac{x-2017}{2019}+\dfrac{x-2019}{2017}=\dfrac{x+6}{2021}\)
\(\Rightarrow\dfrac{x-2017}{2019}-1+\dfrac{x-2019}{2017}-1=\dfrac{x+6}{2021}-2\)
\(\Rightarrow\dfrac{x-2017}{2019}-\dfrac{2019}{2019}+\dfrac{x-2019}{2017}-\dfrac{2017}{2017}=\dfrac{x+6}{2021}-\dfrac{4042}{2021}\)
\(\Rightarrow\dfrac{x-2017-2019}{2019}+\dfrac{x-2019-2017}{2017}=\dfrac{x+6-4042}{2021}\)
\(\Rightarrow\dfrac{x-4036}{2019}+\dfrac{x-4036}{2017}=\dfrac{x-4036}{2021}\)
\(\Rightarrow\dfrac{x-4036}{2021}-\dfrac{x-4036}{2019}-\dfrac{x-4036}{2017}=0\)
\(\Rightarrow\left(x-4036\right)\left(\dfrac{1}{2021}-\dfrac{1}{2019}-\dfrac{1}{2017}\right)=0\)
=> x - 4036 = 0
=> x = 4036
x − 2017/2019 + x−2019/2017 = x+6/2021
=> x − 2017/2019 + x−2019/2017 = x+6/2021
=> x − 2017/2019 − 1 + x − 2019/2017 − 1 = x + 6/2021 − 2
=> x − 2017/2019 − 1 + x − 2019/2017 − 1 = x + 6/2021 − 2
=> x − 2017/2019 − 2019/2019 + x − 2019/2017 − 2017/2017
= x + 6/2021 − 4042/2021
=> x − 2017/2019 − 2019/2019 + x − 2019/2017 − 2017/2017
= x + 6/2021 − 4042/2021
=> x − 2017 − 2019/ 2019 + x − 2019 − 2017/2017
= x + 6 − 4042/2021
=> x − 2017 − 2019/2019 + x − 2019 − 2017/2017 = x + 6 − 4042/2021
=> x − 4036/2019 + x − 4036/2017 = x − 4036/2021
=> x − 4036/2019 + x − 4036/2017 = x − 4036/2021
=> x − 4036/2021 − x − 4036/2019 − x − 4036/2017 = 0
=> x − 4036/2021 − x − 4036/2019 − x − 4036/2017 = 0
=>(x − 4036)(12021 − 12019 − 12017) = 0
=> x - 4036 = 0
=> x = 4036