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NT
0
NT
1
1 tháng 4 2022
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
\(\Leftrightarrow\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1=\dfrac{x-4}{2010}-1+\dfrac{x-5}{2009}-1+\dfrac{x-6}{2008}-1\)
=>x-2014=0
hay x=2014
22 tháng 2 2020
\(\frac{x-2010-2011}{2009}+\frac{x-2009-2011}{2010}+\frac{x-2009-2010}{2011}=3\)
\(\Leftrightarrow\left(\frac{x-2010-2011}{2009}-1\right)+\left(\frac{x-2009-2011}{2010}-1\right)+\left(\frac{x-2009-2010}{2011}-1\right)=0\)
\(\Leftrightarrow\frac{x-6030}{2009}+\frac{x-6030}{2010}+\frac{x-6030}{2011}=0\)
\(\Leftrightarrow\left(x-6030\right)\left(\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}\right)\)
\(\Leftrightarrow x-6030=0\)(vì \(\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}>0\))
\(\Leftrightarrow x=6030\)
Vậy ................
DO
0
TN
2
27 tháng 3 2019
\(\frac{3-x}{2009}-\frac{2-x}{2010}+\frac{1-x}{2011}=-1\)
\(\frac{3-x}{2009}+1-\left(\frac{2-x}{2010}+1\right)+\frac{1-x}{2011}+1=0\)
\(\frac{2012-x}{2009}-\frac{2012-x}{2010}+\frac{2012-x}{2011}=0\)
\(\left(2012-x\right)\left(\frac{1}{2009}-\frac{1}{2010}+\frac{1}{2011}\right)=0\)
Vì \(\frac{1}{2009}-\frac{1}{2010}+\frac{1}{2011}\ne0\)
\(\Rightarrow2012-x=0\)
\(\Rightarrow x=2012\)
LT
1
YN
2 tháng 3 2022
`Answer:`
\(\left(\frac{x+1}{2013}\right)+\left(\frac{x+2}{2012}\right)+\left(\frac{x+3}{2011}\right)=\left(\frac{x+4}{2010}\right)+\left(\frac{x+5}{2009}\right)+\left(\frac{x+6}{2008}\right)\)
\(\Leftrightarrow\frac{x+1}{2013}+1+\frac{x+2}{2012}+1+\frac{x+3}{2011}+1=\frac{x+4}{2010}+1+\frac{x+5}{2009}+1+\frac{x+6}{2008}+1\)
\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}=\frac{x+2014}{2010}+\frac{x+2014}{2009}+\frac{x+2014}{2008}\)
\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}-\frac{x+2014}{2010}-\frac{x+2014}{2009}-\frac{x+2014}{2008}=0\)
\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Rightarrow x+2014=0\)
\(\Leftrightarrow x=-2014\)
PQ
2
ND
4 tháng 3 2018
\(\dfrac{x+1}{2012}+\dfrac{x+2}{2011}=\dfrac{x+3}{2010}+\dfrac{x+4}{2009}\)
\(\Leftrightarrow1+\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}=1+\dfrac{x+3}{2010}+1+\dfrac{x+4}{2009}\) \(\Leftrightarrow\dfrac{x+1+2012}{2012}+\dfrac{x+2+2011}{2011}=\dfrac{x+3+2010}{2010}+\dfrac{x+4+2009}{2009}\) \(\Leftrightarrow\dfrac{x+2013}{2012}+\dfrac{x+2013}{2011}-\dfrac{x+2013}{2010}-\dfrac{x+2013}{2009}=0\) \(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}\right)=0\)
Vì \(\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}\ne0\)
\(\Rightarrow x+2013=0\)
\(\Rightarrow x=-2013\)
Vậy........
4 tháng 3 2018
\(\dfrac{x+1}{2012}+\dfrac{x+2}{2011}=\dfrac{x+3}{2010}+\dfrac{x+4}{2009}\)
\(\Leftrightarrow\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}+1=\dfrac{x+3}{2010}+1+\dfrac{x+4}{2009}+1\)
\(\Leftrightarrow\dfrac{x+2013}{2012}+\dfrac{x+2013}{2011}-\dfrac{x+2013}{2010}-\dfrac{x+2013}{2009}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}\right)=0\)
\(\Leftrightarrow x=-2013\)(vì \(\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}\ne0\))
\(\dfrac{3-x}{2009}-\dfrac{2-x}{2010}+\dfrac{1-x}{2011}=-1\)
=>\(\dfrac{x-3}{2009}+\dfrac{x-2}{2010}-\dfrac{x-1}{2011}=1\)
=>\(\left(\dfrac{x-3}{2009}-1\right)+\left(\dfrac{x-2}{2010}-1\right)-\left(\dfrac{x-1}{2011}-1\right)=0\)
=>\(\dfrac{x-2012}{2009}+\dfrac{x-2012}{2010}-\dfrac{x-2012}{2011}=0\)
=>\(\left(x-2012\right)\left(\dfrac{1}{2009}+\dfrac{1}{2010}-\dfrac{1}{2011}\right)=0\)
=>x-2012=0
=>x=2012