\(\frac{x-1003}{1007}+\frac{x-4}{1003}+\frac{x+2010}{1005}=7\)
K
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NN
0
B
23 tháng 3 2017
x-1009/1001+x-4/1003+x+2010/1005=7
((x-1009/1001)-1))+((x-4/1003)-2)+((x+2010/1005)-4))=0
(x-2010/1001)+(x-2010/1003)+(x-2010/1005)=0
(x-2010)*(1/1001+1/1003+1/1005)=0
okk!!!!!!!!!!!!!!!
KN
30 tháng 4 2019
Đề ???
\(A=\frac{1003+1007+\frac{2010}{113}+\frac{2010}{117}-\frac{1003}{119}-\frac{1007}{119}}{1003+1008+\frac{2011}{113}+\frac{2011}{117}-\frac{1003}{119}-\frac{1008}{119}}\)
\(=\frac{2010+\frac{2010}{113}+\frac{2010}{117}-\frac{2010}{119}}{2011+\frac{2011}{113}+\frac{2011}{117}-\frac{2011}{119}}\)
\(=\frac{2010.\left(1+\frac{1}{113}+\frac{1}{117}-\frac{1}{119}\right)}{2011.\left(1+\frac{1}{113}+\frac{1}{117}-\frac{1}{119}\right)}\)
\(=\frac{2010}{2011}\)
30 tháng 4 2019
\(A=\frac{1003+1007+\frac{2010}{113}+\frac{2010}{117}-\frac{100}{119}-\frac{1007}{119}}{1003+1008+\frac{2011}{113}+\frac{2011}{117}-\frac{1003}{119}-\frac{1008}{119}}\)
\(A=\frac{1003+1008+\frac{2011}{113}+\frac{2011}{117}-\frac{1003}{119}-\frac{1008}{119}}{1003+1008+\frac{2011}{113}+\frac{2011}{117}-\frac{1003}{119}-\frac{1008}{119}}\)+ \(\frac{1+\frac{1}{113}+\frac{1}{117}-\frac{903}{119}-\frac{1}{119}}{1003+1008+\frac{2011}{113}+\frac{2011}{117}-\frac{1003}{119}-\frac{1008}{119}}\)
\(A=1+\frac{1+\frac{1}{113}+\frac{1}{117}-\frac{904}{119}}{2011+\frac{2011}{113}+\frac{2011}{117}-\frac{2011}{119}}\)
\(A=\frac{1+\frac{1}{113}+\frac{1}{117}-\frac{1}{119}-\frac{90.}{119}}{2011+2011.\left(\frac{1}{113}+\frac{1}{117}-\frac{1}{119}\right)}\)
\(A=\frac{\frac{90}{119}}{2010+2011}\)
\(A=\frac{\frac{90}{119}}{4021}\)
SO
23 tháng 3 2020
\(\frac{x-1009}{1001}\)+\(\frac{x-4}{1003}\)+\(\frac{x+2010}{1005}\)=7
⇔\(\frac{x-1009}{1001}\)+\(\frac{x-4}{1003}\)+\(\frac{x+2010}{1005}\)-7=0
⇔\(\left(\frac{x-1009}{1001}-1\right)+\left(\frac{x-4}{1003}-2\right)+\left(\frac{x+2010}{1005}-4\right)=0\)
⇔\(\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)
⇔(x-2010)\(\left(\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}\right)\)=0
⇔x-2010=0
⇔x=2010
Vậy x=2010
DT
23 tháng 3 2020
\(\frac{x-1009}{1001}+\frac{x-4}{1003}+\frac{x+2010}{1005}=7\)
⇔ \(\frac{x-1009}{1001}+\frac{x-4}{1003}+\frac{x+2010}{1005}-7=0\)
⇔\(\left(\frac{x-1009}{1001}-1\right)+\left(\frac{x-4}{1003}-2\right)\)\(+\left(\frac{x+2010}{1005}-4\right)=0\)
⇔\(\frac{x-1009-1001}{1001}+\frac{x-4-2006}{1003}+\)\(\frac{x+2010-4020}{1005}=0\)
⇔\(\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)
⇔\(\left(x-2010\right)\left(\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}\right)=0\)
⇔ \(x-2010=0\left(do\frac{1}{1001}+\frac{1}{1003}+\frac{1}{1005}>0\right)\)
⇔ \(x=2010\)
Vậy S = {2010}
27 tháng 5 2015
\(1-\frac{1003}{1005}=\frac{2}{1005}>\frac{2}{1007}=1-\frac{1005}{1007}\Rightarrow\frac{1003}{1005}
27 tháng 5 2015
ta có : 1-1003/1005=2/1005
1-1005/1007=2/1007
vì 2/1005>2/1007 nên 1003/1005<1005/1007
M
1
26 tháng 1 2021
Ta có: \(\dfrac{x+1006}{1007}+\dfrac{x+1005}{1008}=\dfrac{x+1004}{1009}+\dfrac{x+1003}{1010}\)
\(\Leftrightarrow\dfrac{x+1006}{1007}+1+\dfrac{x+1005}{1008}+1=\dfrac{x+1004}{1009}+1+\dfrac{x+1003}{1010}+1\)
\(\Leftrightarrow\dfrac{x+2013}{1007}+\dfrac{x+2013}{1008}=\dfrac{x+2013}{1009}+\dfrac{x+2013}{1010}\)
\(\Leftrightarrow\dfrac{x+2013}{1007}+\dfrac{x+2013}{1008}-\dfrac{x+2013}{1009}-\dfrac{x+2013}{1010}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{1007}+\dfrac{1}{1008}-\dfrac{1}{1009}-\dfrac{1}{1010}\right)=0\)
mà \(\dfrac{1}{1007}+\dfrac{1}{1008}-\dfrac{1}{1009}-\dfrac{1}{1010}\ne0\)
nên x+2013=0
hay x=-2013
Vậy: S={-2013}
PM
1
NM
Nguyễn Minh Quang
Giáo viên
18 tháng 3 2022
ta có :
\(\frac{x-1009}{1001}-1+\frac{x-4}{1003}-2+\frac{x+2010}{1005}-4=0\)
hay \(\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\Leftrightarrow x-2010=0\)
hay x =2010
Vậy phương trình có nghiệm x = 2010
\(\frac{x-1003}{1007}+\frac{x-4}{1003}+\frac{x+2010}{1005}=7\)
\(\Rightarrow\left(\frac{x-1003}{1007}-1\right)+\left(\frac{x-4}{1003}-1\right)+(\frac{x+2010}{1005}-4)=0\)
\(\Rightarrow\frac{x-2010}{1007}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\)
\(\Rightarrow\left(x-2010\right)\left(\frac{1}{1007}+\frac{1}{1003}+\frac{1}{1005}\right)\)
Vì
\(\frac{1}{1007}+\frac{1}{1003}+\frac{1}{1005}\ne0\Rightarrow X-2010=0\Rightarrow x=2010\)
\(\frac{x-1003}{1007}+\frac{x-4}{1003}+\frac{x+2010}{1005}=7\)
\(\frac{x-1003}{1007}-1+\frac{x-4}{1003}-2+\frac{x+2010}{1005}-4=0\)
\(\frac{x-2010}{1003}+\frac{x-2010}{1005}+\frac{x-2010}{1007}=0\)
\(\left(x-2010\right)\left(\frac{1}{1003}+\frac{1}{1005}+\frac{1}{1007}\right)=0\)
\(\frac{1}{1003}+\frac{1}{1005}+\frac{1}{1007}\ne0\)
\(\Rightarrow x-2010=0\Rightarrow x=2010\)