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\(\frac{x}{2012}+\frac{x-1}{2013}+\frac{x-2}{2014}-\frac{x-3}{2015}=\frac{x-4}{1008}\)
\(\Leftrightarrow\left(\frac{x}{2012}+1\right)+\left(\frac{x-1}{2013}+1\right)+\left(\frac{x-2}{2014}+1\right)-\left(\frac{x-3}{2015}+1\right)=\frac{x-4}{1008}+2\)
\(\Leftrightarrow\frac{x+2012}{2012}+\frac{x+2012}{2013}+\frac{x+2012}{2014}-\frac{x+2012}{2015}=\frac{x-4+1008.2}{1008}\)
\(\Leftrightarrow\frac{x+2012}{2012}+\frac{x+2012}{2013}+\frac{x+2012}{2014}-\frac{x+2012}{2015}=\frac{x+2012}{1008}\)
\(\Leftrightarrow\frac{x+2012}{2012}+\frac{x+2012}{2013}+\frac{x+2012}{2014}-\frac{x+2012}{2015}-\frac{x+2012}{1008}=0\)
\(\Leftrightarrow\left(x+2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{1008}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{1008}\ne0\)
\(\Rightarrow x+2012=0\)\(\Leftrightarrow x=-2012\)
Vậy \(x=-2012\)
=\(\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)-\left(\frac{x-3}{2013}-1\right)-\left(\frac{x-4}{2012}-1\right)\)
=\(\frac{x-2016}{2015}+\frac{x-2016}{2014}-\frac{x-2106}{2013}-\frac{x-2016}{2012}\)
=\(\left(x-2016\right).\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)\)
Mà: \(\frac{1}{2012}>\frac{1}{2015}\) và \(\frac{1}{2014}< \frac{1}{2013}\)
=>\(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\) khác \(0\)
Nên: \(x-2016=0\)
=>\(x=2016\)
\(\frac{x+4}{2012}+\frac{x+3}{2013}=\frac{x+2}{2014}+\frac{x+1}{2015}\)
\(\Rightarrow\frac{x+4}{2012}+1+\frac{x+3}{2013}+1=\frac{x+2}{2014}+1+\frac{x+1}{2015}+1\)
\(\Rightarrow\frac{x+2016}{2012}+\frac{x+2016}{2013}=\frac{x+2016}{2014}+\frac{x+2016}{2015}\)
\(\Rightarrow\frac{x+2016}{2012}+\frac{x+2016}{2013}-\left(\frac{x+2016}{2014}+\frac{x+2016}{2015}\right)=0\)
\(\Rightarrow\left(x+2016\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\ne0\)
\(\Rightarrow x+2016=0\)
\(\Rightarrow x=-2016\)
CÓ: \(\frac{x-1}{2015}+\frac{x-2}{2014}-\frac{x-3}{2013}-\frac{x-4}{2012}=0\)\(0\)
<=>\(\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)-\left(\frac{x-3}{2013}-1\right)-\left(\frac{x-4}{2012}-1\right)=0\)
<=>\(\frac{x-2016}{2015}+\frac{x-2016}{2014}-\frac{x-2016}{2013}-\frac{x-2016}{2012}=0\)
<=>\(\left(x-2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
Do:\(\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)\ne0\)
=>\(x-2016=0\)
<=>\(x=2016\)
Ta có: \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Mà \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
\(\Rightarrow x+2004=0\)
\(\Rightarrow x=-2004\)
Vậy .........................................
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