tim x biết x+2 / 2012 + x+3 / 2011 = x+4 / 2010 + x+5 / 2009
Những câu hỏi liên quan
tim x biet
x + 4/2009 + x + 3 / 2010 = x + 2/2011 + x + 1/2012
Tìm x biết: (x+1/2013) + (x+2/2012) + (x+3/2011) = (x+4/2010) + (x+5/2009) + (x+6/2008)
`Answer:`
\(\left(\frac{x+1}{2013}\right)+\left(\frac{x+2}{2012}\right)+\left(\frac{x+3}{2011}\right)=\left(\frac{x+4}{2010}\right)+\left(\frac{x+5}{2009}\right)+\left(\frac{x+6}{2008}\right)\)
\(\Leftrightarrow\frac{x+1}{2013}+1+\frac{x+2}{2012}+1+\frac{x+3}{2011}+1=\frac{x+4}{2010}+1+\frac{x+5}{2009}+1+\frac{x+6}{2008}+1\)
\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}=\frac{x+2014}{2010}+\frac{x+2014}{2009}+\frac{x+2014}{2008}\)
\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}-\frac{x+2014}{2010}-\frac{x+2014}{2009}-\frac{x+2014}{2008}=0\)
\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Rightarrow x+2014=0\)
\(\Leftrightarrow x=-2014\)
Tim x:
x . (1/2+1/3+1/4+. . .+1/2011+1/2012)
2012/1 +2011/2+2010/3+2009/4+2/2011+1/2012
=1
X+5/2009+x+4/2010=x+3/2011+x+2/2012
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
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x-1 / 2013 + x-2 / 2012 + x-3 / 2011 = x-4 / 2010 + x-5 / 2009 + x-6 / 2008
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
\(\Leftrightarrow\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1=\dfrac{x-4}{2010}-1+\dfrac{x-5}{2009}-1+\dfrac{x-6}{2008}-1\)
=>x-2014=0
hay x=2014
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Tìm x biết: (x+4)/2009+(x+3)/2010=(x+2)/2011+(x+1)/2012
https://i.imgur.com/9wHQzCB.jpg
5-x/2010+4-x/2011+3-x/2012=6-x/2009+1-x/1007
\(5-\frac{x}{2010}+4-\frac{x}{2011}+3-\frac{x}{2012}=6-\frac{x}{2009}+1-\frac{x}{1007}.\)
\(\left(5+4+3\right)-x.\frac{1}{2010}-x.\frac{1}{2011}-x\frac{1}{2012}=\left(6+1\right)-x.\frac{1}{2009}-x\frac{1}{1007}\)
\(12-x.\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)=7-x.\left(\frac{1}{2009}+\frac{1}{1007}\right)\)
\(-x.\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)+x.\left(\frac{1}{2009}+\frac{1}{1007}\right)=7-12\)
\(x.\left(\frac{-1}{2010}-\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2009}+\frac{1}{1007}\right)=-5\)
\(x=\frac{-5}{\frac{-1}{2010}-\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2009}+\frac{1}{1007}}\)
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tìm x biết
\(\frac{x+2}{2012}+\frac{x+3}{2011}+\frac{x+4}{2010}+\frac{x+5}{2009}\)
Xem thêm câu trả lời
Giải phương trình sau:
x/2008+(x+1)/2009+(x+2)/2010+(x+3)/2011+(x+4)/2012=5
\(\frac{x}{2008}+\frac{x+1}{2009}+...+\frac{x+4}{2012}=5\)
\(\Leftrightarrow\left(\frac{x}{2008}-1\right)+\left(\frac{x+1}{2009}-1\right)+...+\left(\frac{x+4}{2012}-1\right)=0\)
\(\Leftrightarrow\frac{x-2008}{2008}+\frac{x-2008}{2009}+...+\frac{x-2008}{2012}=0\)
\(\Leftrightarrow\left(x-2008\right)\left(\frac{1}{2008}+\frac{1}{2009}+..+\frac{1}{2012}\right)=0\)
Mà \(\left(\frac{1}{2008}+\frac{1}{2009}+..+\frac{1}{2012}\right)\ne0\)
Nên \(x-2008=0\)
\(\Leftrightarrow x=2008\)
Vậy : \(x=2008\)
\(\frac{x}{2008}+\frac{x+1}{2009}+\frac{x+2}{2010}+\frac{x+3}{2011}+\frac{x+4}{2012}=5\)
\(\Leftrightarrow\frac{x}{2008}+\frac{x+1}{2009}+\frac{x+2}{2010}+\frac{x+3}{2011}+\frac{x+4}{2012}-5=0\)
\(\Leftrightarrow\left(\frac{x}{2008}-1\right)+\left(\frac{x+1}{2009}-1\right)+\left(\frac{x+2}{2010}-1\right)+\left(\frac{x+3}{2011}-1\right)+\left(\frac{x+4}{2012}-1\right)=0\)
\(\Leftrightarrow\frac{x-2008}{2008}+\frac{x-2008}{2009}+\frac{x-2008}{2010}+\frac{x-2008}{2011}+\frac{x-2008}{2012}=0\)
\(\Leftrightarrow\left(x-2008\right)\left(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)=0\)
Vì \(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\ne0\)
\(\Rightarrow x-2008=0\)\(\Leftrightarrow x=2008\)
Vậy \(x=2008\)