Cho x,y,z > 0 thỏa mãn điều kiện 

C

- Hướng dẫn giải

Đặt $\frac{x\left(y+z-x\right)}{\log \left(x\right)}=\frac{y\left(z+x-y\right)}{\log \left(y\right)}=\frac{z\left(x+y-z\right)}{\log \left(z\right)}=\frac{1}{t}$

Suy ra

$$\begin{gathered} \log \left(x\right)=tx\left(y+z-z\right) \\ \iff y\log \left(x\right)=txy\left(y+z-x\right) \\ \log \left(y\right)=ty\left(z+x-y\right) \\ \iff x\log \left(y\right)=txy\left(z+x-y\right) \end{gathered}$$

Từ đó ta có 

$$\begin{gathered} x\log \left(y\right)+y\log \left(x\right)=2txyz\left(1\right) \\ y\log \left(z\right)+z\log \left(y\right)=2txyz\left(2\right) \\ z\log \left(x\right)+x\log \left(z\right)=2txyz\left(3\right) \end{gathered}$$

Từ (1), (2) và (3) suy ra

$$\begin{gathered} x\log \left(y\right)+y\log \left(x\right)=y\log \left(z\right)+z\log \left(y\right) \\ =z\log \left(x\right)+x\log \left(z\right) \\ \iff \log \left(x^y{y}^x\right)=\log \left(z^y{y}^z\right)=\log \left(z^x{x}^z\right) \\ \Rightarrow x^y{y}^x=z^y{y}^z=z^x{x}^z \end{gathered}$$

Đáp án C