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

Đề sai.

kể ra tử đều là x+2016 hoặc x+2010 thì còn làm được đó

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 \(x=2014;y=2015\)

x+2/2013+x+1/2014=x/2015+x-1/2016

a) \(\left(\left|x-3\right|+2\right)^2+\left|y+3\right|=2007\)

Ta có: \(\left|x-3\right|\ge0\forall x\)

\(\Rightarrow\left(\left|x-3\right|+2\right)^2\ge\left(0+2\right)^2=2^2=4\)

Lại có: \(\left|y+3\right|\ge0\forall y\)

\(\Rightarrow\left(\left|x-3\right|+2\right)^2+\left|y+3\right|\ge4+0=4\)

\(\Rightarrow\left(\left|x-3\right|+2\right)^2+\left|y+3\right|+2007\ge4+2007=2011\)

 \(\Rightarrow P_{MIN}=2011\)

Dấu "=" xảy ra khi \(\Leftrightarrow\orbr{\begin{cases}\left|x-3\right|=0\\\left|y+3\right|=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\y=-3\end{cases}}}\)

Vậy \(P_{MIN}=2011\) tại \(\orbr{\begin{cases}x=3\\y=-3\end{cases}}\)

\(x=\frac{-2014}{-2013}=\frac{2014}{2013}>1\)

\(y=\frac{-2015}{-2016}=\frac{2015}{2016}