\(\frac{x+2007}{2019}\)+\(\frac{x+2008}{2020}\)=\(\frac{x+2009}{2021}\)+\(\frac{x+2010}{2022}\)
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BM
25 tháng 8 2019
ko ghi lại đề
ta thấy : 2019 - 1 = 2018
2020 - 2 = 2018
2021 - 3 = 2018
2022 - 4 = 2018
=> x = 2018
thử lại :
2018+1/2019 + 2018+2/2020 = 2018+3/2021 + 2018+4/2022
= 1 + 1 = 1 + 1
2 = 2
NH
16 tháng 7 2019
\(\frac{x+4}{2019}+\frac{x+3}{2020}=\frac{x+2}{2021}+\frac{x+1}{2020}\)
\(\Leftrightarrow(\frac{x+4}{2019}+1)+(\frac{x+3}{2020}+1)=(\frac{x+2}{2021}+1)+(\frac{x+1}{2022}+1)\)
\(\Leftrightarrow\frac{x+2023}{2019}+\frac{x+2023}{2020}=\frac{x+2023}{2021}+\frac{x+2023}{2022}\)
\(\Leftrightarrow\frac{x+2023}{2019}+\frac{x+2023}{2020}-\frac{x+2023}{2021}-\frac{x+2023}{2022}=0\)
\(\Leftrightarrow\left(x+2023\right)\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}-\frac{1}{2020}\right)=0\)
\(\Leftrightarrow x+2023=0\)
\(\Leftrightarrow x=-2023\)
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
NN
30 tháng 7 2019
\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}-\frac{x+3}{2009}\)
\(\Leftrightarrow\)\(\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}+\frac{x+3}{2009}=0\)
\(\Leftrightarrow\) \(\left(x+3\right).\left(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\right)=0\)
\(\Leftrightarrow\) \(x+3=0\) ( Vì \(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\ne0\) )
\(\Leftrightarrow\) \(x=-3\)
Vậy x = -3
30 tháng 7 2019
\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}-\frac{x+3}{2009}\)
\(\Rightarrow\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}+\frac{x+3}{2009}=0\)
\(\Rightarrow\left(x+3\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\right)=0\)
\(\Rightarrow x+3=0\Leftrightarrow x=-3\)
HT
1 tháng 9 2015
\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}+\frac{x+3}{2009}\)
=> \(\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}-\frac{x+3}{2009}=0\)
=> \(\left(x+3\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)=0\)
=> x + 3 = 0
=> x = 0 - 3
=> x = -3
26 tháng 9 2017
Cho x,y là các số nguyên dương, chứng minh rằng:
\(1< \frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}< 2\)
NQ
30 tháng 7 2015
\(\left(\frac{x+4}{2007}+1\right)+\left(\frac{x+3}{2008}+1\right)=\left(\frac{x+2}{2009}+1\right)+\left(\frac{x+1}{2010}+1\right)\)
\(\left(\frac{x+2011}{2007}\right)+\left(\frac{x+2011}{2008}\right)=\left(\frac{x+2011}{2009}\right)+\left(\frac{x+2011}{2010}\right)\)
\(\frac{x+2011}{2007}+\frac{x+2011}{2008}-\frac{x+2011}{2009}-\frac{x+2011}{2010}=0\)
\(\left(x+2011\right).\left(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)=0\)
Vì \(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\)khác 0 (các số hạng ko bằng nhau)
\(\Leftrightarrow\)\(x+2011=0\)
\(\Rightarrow x=0-2011\)
\(\Rightarrow x=-2011\)
6 tháng 2 2018
Ta có :
\(\frac{x+1}{2012}+\frac{x+2}{2011}+\frac{x+3}{2010}=\frac{x+4}{2009}+\frac{x+5}{2008}+\frac{x+6}{2007}\)
\(\left(\frac{x+1}{2012}+1\right)+\left(\frac{x+2}{2011}+1\right)+\left(\frac{x+3}{2010}+1\right)=\left(\frac{x+4}{2009}+1\right)+\left(\frac{x+5}{2008}+1\right)+\left(\frac{x+6}{2007}+1\right)\)
\(\Leftrightarrow\)\(\frac{x+2013}{2012}+\frac{x+2013}{2011}+\frac{x+2013}{2010}=\frac{x+2013}{2009}+\frac{x+2013}{2008}+\frac{x+2013}{2007}\)
\(\Leftrightarrow\)\(\left(x+2013\right).\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}\right)=\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\right)\)
\(\Leftrightarrow\)\(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}=\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\)\(\left(1\right)\)
Mà \(\frac{1}{2012}< \frac{1}{2009}\)\(;\)\(\frac{1}{2011}< \frac{1}{2008}\)\(;\)\(\frac{1}{2010}< \frac{1}{2007}\)
\(\Rightarrow\)\(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}< \frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\)\(\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)suy ra không có giá trị nào của \(x\)thoả mãn đề bài
Vậy không có gía trị nào của \(x\)hay \(x\in\left\{\varnothing\right\}\)
\(\frac{x+2007}{2019}+\frac{x+2008}{2020}=\frac{x+2009}{2021}+\frac{x+2010}{2022}\)
\(\left(\frac{x+2007}{2019}-1\right)+\left(\frac{x+2008}{2020}-1\right)=\left(\frac{x+2009}{2021}-1\right)+\left(\frac{x+2010}{2022}-1\right)\)
\(\frac{x-12}{2019}+\frac{x-12}{2020}=\frac{x-12}{2021}+\frac{x-12}{2022}\)
\(\frac{x-12}{2019}+\frac{x-12}{2020}-\frac{x-12}{2021}-\frac{x-12}{2022}=0\)
\(\left(x-12\right)\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}-\frac{1}{2022}\right)=0\)
\(\Rightarrow x-12=0\left(do\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}-\frac{1}{2022}\ne0\right)\)
\(x=0+12\)
\(x=12\)
\(\frac{x+2007}{2019}+\frac{x+2008}{2020}=\frac{x+2009}{2021}+\frac{x+2010}{2022}\)
<=> \(\left(\frac{x+2007}{2019}-1\right)+\left(\frac{x+2008}{2020}-1\right)=\left(\frac{x+2009}{2021}-1\right)+\left(\frac{x+2010}{2022}-1\right)\)
<=> \(\frac{x-12}{2019}+\frac{x-12}{2020}=\frac{x-12}{2021}+\frac{x-12}{2022}\)
<=> \(\frac{x-12}{2019}+\frac{x-12}{2020}-\frac{x-12}{2021}-\frac{x-12}{2022}=0\)
<=> \(\left(x-12\right)\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}-\frac{1}{2022}\right)=0\)
Dễ thấy \(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}-\frac{1}{2022}\ne0\)
nên pt <=> x - 12 = 0 <=> x = 12
Vậy ...