Ta tính vế sau:

1+1/3-1+1/3=0

Vì đây là phép nhân nên nếu có một vế bằng 0 thì vế sau cx bằng 0

Ta có:\(\left|x-2013\right|+\left|x-2014\right|+\left|y-2015\right|+\left|x-2016\right|\)

\(=\left|x-2013\right|+\left|2016-x\right|+\left|x-2014\right|+\left|y-2015\right|\)

\(\ge\left|x-2013+2016-x\right|+\left|x-2014\right|+\left|y-2015\right|\)

\(=3+\left|x-2014\right|+\left|y-2015\right|\)

\(\ge3+0+0=3\)

Mà \(\left|x-2013\right|+\left|x-2014\right|+\left|y-2015\right|+\left|x-2016\right|=3\)

\(\Rightarrow\) Dấu "=" xảy ra khi và chỉ khi:

\(\hept{\begin{cases}\left(x-2013\right)\left(2016-x\right)\ge0\\\left|x-2014\right|=0\\\left|y-2015\right|=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2013\le x\le2016\left(1\right)\\x=2014\left(2\right)\\y=2015\end{cases}}\)

Dễ thấy \(\left(2\right)\) thỏa mãn \(\left(1\right)\) nên \(x=2014;y=2015\)

( 2013 x 2014 +2014 x 2015 + 2015 x 2016 ) x ( 1 + 1/3 - 1 - 1/3 )

= ( 2013 x 2014 + 2014 x 2015 + 2015 x 2016 ) x 0

= 0

( 2013 x 2014 + 2014 x 2015 + 2015 x 2016) x ( 1 + 1/3 - 4/3)

=( 2013 x 2014 + 2014 x 2015 + 2015 x 2016) x ( 4/3 - 4/3)

=( 2013 x 2014 + 2014 x 2015 + 2015 x 2016) x 0

=0

Ta có: \(\left(2013\cdot2014+2014\cdot2015+2015\cdot2016\right)\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)

\(=\left(2013\cdot2014+2014\cdot2015+2015\cdot2016\right)\left(\dfrac{3}{3}+\dfrac{1}{3}-\dfrac{4}{3}\right)\)

=0

Min D = 2 <=> x= 2014

\(\dfrac{x+2016}{2013}+\dfrac{x+2010}{2014}+\dfrac{x+2010}{2015}+\dfrac{x+2010}{2016}+\dfrac{x+2010}{2015}+\dfrac{x+2016}{2018}\)

Đề sai.