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\(\frac{x+1006}{1007}+\frac{x+1005}{1008}=\frac{x+1004}{1009}+\frac{x+1003}{1010}\)
\(\Rightarrow\left(\frac{x+1006}{1007}+1\right)+\left(\frac{x+1005}{1008}+1\right)=\left(\frac{x+1004}{1009}+1\right)+\left(\frac{x+1003}{1010}+1\right)\)
\(\Rightarrow\frac{x+2013}{1007}+\frac{x+2013}{1008}=\frac{x+2013}{1009}+\frac{x+2013}{1010}\)
\(\Rightarrow\frac{x+2013}{1007}+\frac{x+2013}{1008}-\frac{x+2013}{1009}-\frac{x+2013}{1010}=0\)
\(\Rightarrow\left(x+2013\right)\left(\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1009}-\frac{1}{1010}\right)=0\)
Mà \(\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1009}-\frac{1}{1010}\ne0\)
\(\Rightarrow x+2013=0\)
\(\Rightarrow x=-2013\)
Vậy x = -2013
a, \(\frac{x+1006}{1000}+\frac{x+1007}{999}+\frac{x+1008}{998}+\frac{x+1009}{997}+\frac{x+2022}{4}=0\)
\(\Leftrightarrow\frac{x+1006}{1000}+1+\frac{x+1007}{999}+1+\frac{x+1008}{998}+1+\frac{x+1009}{997}+1+\frac{x+2022}{4}-4=0\)
\(\Leftrightarrow\frac{x+2006}{1000}+\frac{x+2006}{999}+\frac{x+2006}{998}+\frac{x+2006}{997}+\frac{x+2006}{4}=0\)
\(\Leftrightarrow\left(x+2006\right)\left(\frac{1}{1000}+\frac{1}{999}+\frac{1}{998}+\frac{1}{997}+\frac{1}{4}\right)=0\)
Mà \(\frac{1}{1000}+\frac{1}{999}+\frac{1}{998}+\frac{1}{997}+\frac{1}{4}\ne0\)
\(\Rightarrow x+2006=0\Leftrightarrow x=-2006\)
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}
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!!!!!!!!!!!!!!!
Lời giải:
Tập xác định của phương trình
Sử dụng tính chất tỉ lệ thức, có thể biến đổi phương trình như sau
Lời giải thu được
\(\frac{x+2012}{2}+\frac{x+2010}{3}+\frac{x+2011}{5}=\frac{x}{1008}+\frac{x-2}{1009}+\frac{x+1}{2015}\)
\(\Leftrightarrow\frac{x+2012}{2}+\frac{x+2010}{3}+\frac{x+2011}{5}-\frac{x}{1008}-\frac{x-2}{1009}-\frac{x+1}{2015}=0\)
\(\Leftrightarrow\frac{x+2012}{2}+2+\frac{x+2010}{3}+2+\frac{x+2011}{5}+1-\frac{x}{1008}-2-\frac{x-2}{1009}-2-\frac{x+1}{2015}-1=0\)
\(\Leftrightarrow\frac{x+2016}{2}+\frac{x+2016}{3}+\frac{x+2016}{5}-\frac{x+2016}{1008}-\frac{x+2016}{1009}-\frac{x+2016}{2015}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}-\frac{1}{1008}-\frac{1}{1009}-\frac{1}{2015}\right)=0\)
Mà \(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}-\frac{1}{1008}-\frac{1}{1009}-\frac{1}{2015}\ne0\)
\(\Leftrightarrow x+2016=0\)
\(\Leftrightarrow x=-2016\)
Vậy tập nghiệm của phương trình là \(S=\left\{-2016\right\}\)
\(\frac{x-1009}{1010}+\frac{x-1007}{1012}=\frac{x-1010}{1009}+\frac{x-1012}{1007}\)
\(\Rightarrow(\frac{x-1009}{1010}-1)+\left(\frac{x-1007}{1012}-1\right)=\left(\frac{x-1010}{1009}-1\right)+\left(\frac{x-1012}{1007}-1\right)\)
\(\Rightarrow\frac{x-2019}{1010}+\frac{x-2019}{1012}-\frac{x-2019}{1009}-\frac{x-2019}{1007}\)
\(\Rightarrow\left(x-2019\right)\left(\frac{1}{1010}+\frac{1}{1012}-\frac{1}{1009}-\frac{1}{1007}\right)=0\)
Ta có
\(\frac{1}{1010}+\frac{1}{1012}-\frac{1}{1009}-\frac{1}{1007}\ne0\Rightarrow x-2019=0\Rightarrow x=2019\)
\(\frac{x-1009}{1010}+\frac{x-1007}{1012}=\frac{x-1010}{1009}+\frac{x-1012}{1007}\)
\(\frac{x-1009}{1010}-1+\frac{x-1007}{1012}-1=\frac{x-1010}{1009}-1+\frac{x-1012}{1007}\)\(\frac{x-2019}{1010}+\frac{x-2019}{1012}-\frac{x-2019}{1009}-\frac{x-2019}{1007}=0\)
\(\left(x-2019\right)\left(\frac{1}{1010}+\frac{1}{1012}-\frac{1}{1009}-\frac{1}{1007}\right)=0\)
1/1010 + 1/1012 - 1/1009 - 1/1007 khác 0
=> x - 2019 =0 => x = 2019