\(x-2014-\frac{2015}{2013}+x-2013-\frac{2015}{2014}+x-2014-\frac{2013}{2015}=3\)

\(\Rightarrow\left(x+x+x\right)+\left(-2014-2014\right)-2013-\frac{2015}{2013}-\frac{2015}{2014}-\frac{2013}{2015}=3\)

\(3x-2013-\frac{2015}{2013}-\frac{2015}{2014}-\frac{2013}{2015}=3\)

\(3x=3+2013+\frac{2015}{2013}+\frac{2015}{2014}+\frac{2013}{2015}\)

bạn ơi bài này số lớn quá bạn sử dungjmays tính rồi tự tính nhé

Đáp án của bạn Hoàng Đình Đại sai rùi nhưng dù sao cx cảm ơn nhiều

\(\frac{x+4}{2012}+\frac{x+3}{2013}=\frac{x+2}{2014}+\frac{x+1}{2015}\)

=> \(\frac{x+4}{2012}+1+\frac{x+3}{2013}+1=\frac{x+2}{2014}+1+\frac{x+1}{2015}+1\)

=> \(\frac{x+2016}{2012}+\frac{x+2016}{2013}=\frac{x+2016}{2014}+\frac{x+2016}{2015}\)

=> \(\frac{x+2016}{2012}+\frac{x+2016}{2013}-\frac{x+2016}{2014}-\frac{x+2016}{2015}=0\)

=> \(\left(x+2016\right).\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)

Vì \(\frac{1}{2012}>\frac{1}{2014};\frac{1}{2013}>\frac{1}{2015}\)

=> \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\ne0\)

=> \(x+2016=0\)

=> \(x=-2016\)

\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}-\dfrac{x-3}{2014}=\dfrac{x-4}{2013}\)

\(\Leftrightarrow\dfrac{x-1}{2016}+\dfrac{x-2}{2015}=\dfrac{x-4}{2013}+\dfrac{x-3}{2014}\)

\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)=\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-3}{2014}-1\right)\)

\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{2013}+\dfrac{x-2017}{2014}\)

\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}=0\)

\(\Leftrightarrow x-2017.\left(\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)

\(\text{Mà }\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2103}\ne0\Rightarrow x-2017=0\)

\(\Leftrightarrow x=2017\)         \(\text{Vậy }x=2017\)