Cho 0 < x; y; z

C

- Hướng dẫn giải

$\forall a>0;a\ne 1$ ta có ${\log }_a\left(xyz\right)=0$

$\Rightarrow {\log }_a\left(x\right)+{\log }_a\left(y\right)+{\log }_a\left(z\right)$ = 0

Đặt m = ${\log }_a\left(x\right)$; n = ${\log }_a\left(y\right)$; p = ${\log }_a\left(z\right)$ nên m + n + p = 0.

Theo tính chất của lôgarit, ta viết lại biểu thức S như sau:

$$\begin{gathered} S=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right) \\ \left(\frac{p}{m-n}+\frac{m}{n-p}+\frac{n}{p-m}\right) \end{gathered}$$

Ta có 

$$\begin{gathered} \frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n} \\ =\frac{mn\left(m-n\right)+np\left(n-p\right)}{mnp} \\ \frac{pm\left(p-m\right)}{1} \\ =-\frac{\left(m-n\right)\left(n-p\right)\left(p-m\right)}{mnp} \\ \frac{p}{m-n}+\frac{m}{n-p}+\frac{n}{p-m} \\ =\frac{p\left(n-p\right)\left(p-m\right)}{\left(m-n\right)} \\ \frac{m\left(m-n\right)\left(p-m\right)}{\left(n-p\right)} \\ \frac{n\left(m-n\right)\left(n-p\right)}{\left(p-m\right)} \\ =\frac{mn\left(m+n\right)+np\left(n+p\right)}{\left(m-n\right)} \\ +\frac{\left(p+m\right)-6mnp}{\left(n-p\right)} \\ -\frac{\left(m^3+n^3+p^3-3mnp\right)}{\left(p-m\right)} \\ =\frac{mn\left(-p\right)+np\left(-m\right)}{\left(m-n\right)\left(n-p\right)} \\ +\frac{pm\left(-n\right)-6mnp}{\left(p-m\right)} \\ =\frac{-9mnp}{\left(m-n\right)\left(n-p\right)\left(p-m\right)} \end{gathered}$$

Vậy S = 9

Đáp án C