Vì \(\hept{\begin{cases}\left(x-2\right)^{2012}\ge0;\forall x,y\\\left|y^2-9\right|^{2014}\ge0;\forall x,y\end{cases}}\)\(\Rightarrow\left(x-2\right)^{2012}+\left|y^2-9\right|^{2014}\ge0;\forall x,y\)

Do đó \(\left(x-2\right)^{2012}+\left|y^2-9\right|^{2014}=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^{2012}=0\\\left|y^2-9\right|^{2014}=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=2\\y=\pm3\end{cases}}\)

Vậy \(\left(x,y\right)=\left\{\left(2;3\right);\left(2;-3\right)\right\}\)

vì (x-2)^2012 \(\ge\)0 với mọi x   (1)

 \(|y^2-9|^{2014}\ge0\) với mọi y    (2) 

Mà (x-2)^2012 +\(|y^2-9|^{2014}=0\) (3)

Từ (1), (2), (3) suy ra (x-2)^2012 =0 và \(|y^2-9|^{2014}=0\)

suy ra x=2 và y^2=9

Suy ra x=2 và y=\(\pm\)3

Vì \(\left(x-2\right)^{2012}\ge0\forall x\\ \left|y^2-9\right|^{2014}\ge0\forall y\)

Nên (x-2)^2012+∣y^2−9∣^2014=0

\(\Leftrightarrow\begin{cases}\left(x-2\right)^{2012}=0\\\left|y^2-9\right|^{2014}=0\end{cases}\)

\(\Leftrightarrow\begin{cases}x-2=0\\y^2-9=0\end{cases}\)

\(\Leftrightarrow\begin{cases}x=2\\y^2=9\end{cases}\)

\(\Leftrightarrow\begin{cases}x=2\\y=\pm3\end{cases}\)

<=> \(\frac{\left(x+2014\right)}{2011}+1+\frac{\left(x+2013\right)}{2012}+1=\frac{\left(x+2012\right)}{2013}+1+\frac{\left(x+2011\right)}{2014}+1\)

\(\Rightarrow\frac{\left(x+4025\right)}{2011}+\frac{\left(x+4025\right)}{2012}=\frac{\left(x+4025\right)}{2013}+\frac{\left(x+4025\right)}{2014}\)

=> \(\frac{\left(x+4025\right)}{2011}+\frac{\left(x+4025\right)}{2012}-\frac{\left(x+4025\right)}{2013}-\frac{\left(x+4025\right)}{2014}=0\)

=> \(\left(x+4025\right)\left\lbrack\left(\frac{1}{2011}+\frac{1}{2012}\right)-\left(\frac{1}{2013}+\frac{1}{2014}\right)\right\rbrack=0\)

vì \(\left(\frac{1}{2011}+\frac{1}{2012}\right)>\left(\frac{1}{2013}+\frac{1}{2014}\right)\)

=> \(\left\lbrack\left(\frac{1}{2011}+\frac{1}{2012}\right)-\left(\frac{1}{2013}+\frac{1}{2014}\right)\right\rbrack>0\) hay ≠0

=> \(x+4025=0\)

\(x=-4025\)

\(x=2\)

\(y=3\)

\(\Rightarrow x\cdot y=2\cdot3=6\)

x=2

y=3

\(\Rightarrow x.y=2.3=6\)NHA  BAN

Ta có: \(x^{2014}+x^{2012}+x^{2010}+...+x^2+1=0\)

=>\(x^{2014}+x^{2012}+x^{2010}+...+x^2=0-1=\left(-1\right)\)

Vô lí => X không có giá trị .

Có\(\left|x-2010\right|+\left|x-2012\right|+\left|x-2014\right|\ge\left|x-2010+2014-x\right|+\left|x-2012\right|\ge2\)

mà\(\left|x-2010\right|+\left|x-2012\right|+\left|x-2014\right|=2\)

dấu "=' \(\Leftrightarrow\left\{{}\begin{matrix}x-2012=0\\2010\le x\le2014\end{matrix}\right.\)\(\Rightarrow x=2012\)

Thay vào thì \(|x-2010|+|x-2012|+|x-2014|=4\) (Vô lý)

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

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

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

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

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

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

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

=> \(x+2016=0\)

=> \(x=-2016\)