\(A=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)\(=\left(\frac{x+y}{y}\right)\left(\frac{y+z}{z}\right)\left(\frac{z+x}{x}\right)\)

Xét 2 TH

+> Nếu \(x+y+z=0\)

=> \(\hept{\begin{cases}x+y=-z\\y+z=-x\\x+z=-y\end{cases}}\)

=> \(A=\left(-\frac{z}{y}\right)\left(-\frac{x}{z}\right)\left(-\frac{y}{x}\right)=-1\)

+> Nếu \(x+y+z\ne0\)

\(\frac{x+y+2013z}{z}=\frac{y+z+2013x}{x}=\frac{x+z+2013y}{y}\)

=> \(\frac{x+y}{z}+2013=\frac{y+z}{x}+2013=\frac{z+x}{y}+2013\)

=>\(\frac{x+y}{z}=\frac{y+z}{x}=\frac{z+x}{y}\)\(=\frac{x+y+y+z+z+x}{x+y+z}=2\)

=> \(\hept{\begin{cases}x+y=2z\\y+z=2x\\z+x=2y\end{cases}}\)

=> A = 2.2.2=8

Ta có :

\(A=\frac{x+y+2013z}{z}=\frac{y+z+2013x}{x}=\frac{x+z+2013}{y}\)

\(\Leftrightarrow A=\frac{x+y}{z}+2013=\frac{y+z}{x}+2013=\frac{x+z}{y}+2013=2015\)( Chỗ này áp dụng Tc của dãy tỉ số bằng nhau là ra )

\(\Leftrightarrow\frac{x+y}{z}=\frac{y+z}{x}=\frac{x+z}{y}=2\)

\(\Rightarrow\hept{\begin{cases}x+y=2z\\y+z=2x\\x+z=2y\end{cases}}\)

Thay vào ta có :

\(\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)

\(=\left(\frac{x+y}{y}\right)\left(\frac{y+z}{z}\right)\left(\frac{x+z}{x}\right)\)

\(=\frac{2z.2x.2y}{xyz}=\frac{8xyz}{xyz}=8\)

Vậy ...........

x,y,z=0

Đặt \(\frac{x}{2011}=\frac{y}{2012}=\frac{z}{2013}=k\)

\(\Rightarrow\hept{\begin{cases}x=2011k\\y=2012k\\z=2013k\end{cases}}\)

+) Ta có : \(\frac{2012z-2013y}{2011}=\frac{2012.2013k-2013.2012k}{2011}=0\)

\(\frac{2013x-2011z}{2012}=\frac{2013.2011k-2011.2013k}{2012}=0\)

\(\frac{2011y-2012x}{2013}=\frac{2011.2012k-2012.2011k}{2013}=0\)

Do đó : \(\frac{2012z-2013y}{2011}=\frac{2013x-2011z}{2012}=\frac{2011y-2012x}{2013}\left(=0\right)\) ( đpcm )

Đặt \(\frac{x}{2011}=\frac{y}{2012}=\frac{z}{2013}=k\Rightarrow\hept{\begin{cases}x=2011k\\y=2012k\\z=2013k\end{cases}}\)

\(\frac{2012z-2013y}{2011}=\frac{2012\cdot2013k-2013k\cdot2012}{2011}=\frac{0}{2011}=0\)(1)

\(\frac{2013x-2011z}{2012}=\frac{2013\cdot2011k-2011\cdot2013k}{2012}=\frac{0}{2012}=0\)(2)

\(\frac{2011y-2012x}{2013}=\frac{2011\cdot2012k-2012\cdot2011k}{2013}=\frac{0}{2013}=0\)(3)

Từ (1) , (2) và (3) => đpcm

Áp dụng tính chất của dãy tỉ số = nhau ta có:

\(\frac{x}{y}=\frac{y}{z}=\frac{z}{x}=\frac{x+y+z}{y+z+x}=1\)

=> x = y = z

Ta có: \(A=\frac{2013x^2+y^2+z^2}{x^2+2013y^2+z^2}=\frac{2013x^2+x^2+x^2}{x^2+2013x^2+x^2}=\frac{2015x^2}{2015x^2}=1\)

\(DK:\hept{\begin{cases}x>0&y>\frac{2012}{2013}&\end{cases}}\)

HPT

\(\text{ }\Leftrightarrow\hept{\begin{cases}2013\sqrt{2013y-2012}=\frac{2013}{x}\left(1\right)\\y^2+2012=\frac{2013}{x}\left(2\right)\end{cases}}\)

\(\left(1\right),\left(2\right)\Rightarrow y^2-2013\sqrt{2013y-2012}+2012=0\)

\(\Leftrightarrow\left(y^2-1\right)-2013\left(\sqrt{2013y-2012}-1\right)=0\)

\(\Leftrightarrow\left(y+1\right)\left(y-1\right)-\frac{2013^2\left(y-1\right)}{\sqrt{2013y-2012}+1}=0\)

\(\Leftrightarrow\left(y-1\right)\left(y+1-\frac{2013^2}{\sqrt{2013y-2012}+1}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=1\\y+1-\frac{2013^2}{\sqrt{2013y-2012}+1}=0\end{cases}}\)

Cai PT thu to ay vo nghiem nhung biet chung minh :)

\(\Rightarrow x=1\)

Vay nghiem cua HPT la \(\left(x;y\right)=\left(1;1\right)\)