\(\Rightarrow x^{2014}+y^{2014}-2\left(x^{2013}+y^{2013}\right)+x^{2012}+y^{2012}=0\)

\(\Leftrightarrow x^{2012}.\left(x-1\right)^2+y^{2012}.\left(y-1\right)^2=0\)

\(\Rightarrow x=1;y=1\)

\(\Rightarrow P=2\)

cai gi

Nếu \(\frac{x}{2013}=\frac{y}{2014}=\frac{z}{2015}\Rightarrow x=y=z=0\)

Vậy     \(T=\frac{\left(x-z\right)^2}{\left(x-y\right)^2.\left(y-z\right)}=\frac{0^2}{0^2.0}\)   mà phân số được viết dưới dạng \(\frac{a}{b}\) với a thuộc Z và b khác 0

\(\Rightarrow\)T không có giá trị thỏa mãn

Ta có:

\(x^3+y^3+z^3=3xyz\)

nên  \(x^3+y^3+z^3-3xyz=0\)

\(\Leftrightarrow\left(x^3+y^3\right)+z^3-3xyz=0\)

\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)

\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2+\left(x+y\right).z+z^2\right]-3xy\left(x+y+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2-3xy\right]=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)

\(\Leftrightarrow\frac{1}{2}\left(x+y+z\right)\left(2x^2+2y^2+2z^2-2xy-2xz-2yz\right)=0\)

\(\Leftrightarrow\frac{1}{2}\left(x+y+z\right)\left[\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)\right]=0\)

\(\Leftrightarrow\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)

\(\Leftrightarrow^{x+y+z=0}_{x=y=z}\)

Do đó:

\(M=\left(2-\frac{x}{y}\right)^{2013}+\left(3-\frac{2x}{z}\right)^{2014}+\left(4-\frac{3z}{x}\right)^{2015}\)

\(=\left(2-\frac{y}{y}\right)^{2013}+\left(3-\frac{2z}{z}\right)^{2014}+\left(4-\frac{3x}{x}\right)^{2015}\)

\(=\left(2-1\right)^{2013}+\left(3-2\right)^{2014}+\left(4-3\right)^{2015}\)

\(M=1^{2013}+1^{2014}+1^{2015}=1+1+1=3\)

                                                    ----------------------------------------------------

=4050151.001

2012×2013+2011/2014×2013-2015=2012×2013+2011/(2012+2)×2013-2015=2012×2013+2011/2012×2013+2×2015=2012×2013+2011/2012×2013+4026-2015=2012×2013+2011/2012×2013+2011=1

ai tick cho thêm 20 cái tròn 200 điểm lun

\(\frac{x}{2013}=\frac{y}{2014}=\frac{z}{2015}\Rightarrow\frac{2014.2015.x}{2013.2014.2015}=\)\(\frac{y.2013.2015}{2013.2014.2015}=\frac{2013.2014.z}{2013.2014.2015}\)

\(\Rightarrow2014.2015.x=y.2013.2015=z.2013.2014\)

\(\Rightarrow x=2013;y=2014;z=2015\)

Đến đây bạn tự thay vào rồi tính nhé!

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

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

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

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

\(x+2016=0\)

\(x=-2016\)

\(A=\left|x-2011\right|+\left|x-2012\right|+\left|x-2013\right|+\left|x-2014\right|+\left|x-2015\right|\)

\(=\left(\left|x-2011\right|+\left|x-2015\right|\right)+\left(\left|x-2012\right|+\left|x-2014\right|\right)+\left|x-2013\right|\)

Đặt \(B=\left|x-2011\right|+\left|x-2015\right|\)

\(=\left|x-2011\right|+\left|2015-x\right|\ge\left|x-2011+2015-x\right|=4\left(1\right)\)

Dấu"=" xảy ra \(\Leftrightarrow\left(x-2011\right)\left(2015-x\right)\ge0\)

\(\Leftrightarrow\hept{\begin{cases}x-2011\ge0\\2015-x\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x-2011< 0\\2015-x< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge2011\\x\le2015\end{cases}}\)hoặc \(\hept{\begin{cases}x< 2011\\x>2015\end{cases}\left(loai\right)}\)

\(\Leftrightarrow2011\le x\le2015\)

Đặt \(C=\left|x-2012\right|+\left|x-2014\right|\)

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

Dấu"="xảy ra \(\Leftrightarrow\left(x-2012\right)\left(2014-x\right)\ge0\)

\(\Leftrightarrow\hept{\begin{cases}x-2012\ge0\\2014-x\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x-2012< 0\\2014-x< 0\end{cases}}\) 

\(\Leftrightarrow\hept{\begin{cases}x\ge2012\\x\le2014\end{cases}}\)hoặc\(\hept{\begin{cases}x< 2012\\x>2014\end{cases}\left(loai\right)}\)

\(\Leftrightarrow2012\le x\le2014\)

Ta có: \(\left|x-2013\right|\ge0;\forall x\left(3\right)\)

Dấu"="Xảy ra \(\Leftrightarrow\left|x-2013\right|=0\)

                      \(\Leftrightarrow x=2013\)

Từ (1),(2) và (3) \(\Rightarrow B+C+\left|x-2013\right|\ge6\)

Hay \(A\ge6\)

Dấu"="xảy ra \(\Leftrightarrow\hept{\begin{cases}2011\le x\le2015\\2012\le x\le2014\\x=2013\end{cases}}\)\(\Leftrightarrow x=2013\)

Vậy \(A_{min}=6\Leftrightarrow x=2013\)