tim x biet
x + 4/2009 + x + 3 / 2010 = x + 2/2011 + x + 1/2012
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DT
0
PT
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
PQ
1
BS
0
26 tháng 8 2016
x+5/2009 + x+4/2010 = x+3/2011 + x+2/2012
=> 1 + x+5/2009 + 1 + x+4/2000 = 1 + x+3/2011 + 1 + x+2/2012
=> x+2014/2009 + x+2014/2000 = x+2004/2011 + x+2014/2012
=> x+2014/2009 + x+2014/2000 - x+2014/2011 - x+2014/2012 = 0
=> (x+2014).(1/2009 + 1/2010 - 1/2011 - 1/2012) = 0
Do 1/2009 > 1/2011; 1/2010 > 1/2012
=> 1/2009 + 1/2010 - 1/2011 - 1/2012 khác 0
=> x + 2014 = 0
=> x = -2014
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 \(\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}\ne0\))
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