Đáp án:
$A = 0$
Giải thích các bước giải:
Đặt $2^x = 5^y = 10^{-z}= k$
$\to \begin{cases}x = \log_2k\\y =\log_5k\\z = -\log_{10}k\end{cases}$
$\to A = xy + yz + zx =\log_2k.\log_5k - \log_5k.\log_{10}k - \log_{10}k.\log_2k$
$\to A = \dfrac{1}{\log_k2.\log_k5} -\dfrac{1}{\log_k5.\log_k10} -\dfrac{1}{\log_k10.\log_k2}$
$\to A =\dfrac{\log_k10 - \log_k2 - \log_k5}{\log_k2\log_k5\log_k10}$
$\to A =\dfrac{\log_k1}{\log_k2\log_k5\log_k10}$
$\to A =\dfrac{0}{\log_k2\log_k5\log_k10}= 0$