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1. $\frac{1}{(\log_{a} b c) + 1} +$ $\frac{1}{(\log_{b} c a) + 1} +$ $\frac{1}{(\log_{c} a b) + 1}$ is equal to -

A. 1

B. $\frac{3}{2}$

C. 2

D. 3

Solution:

\begin{aligned} Given Expression \\ = \frac{1}{\log_{a} b c + \log_{a} a} + \frac{1}{\log_{b} c a + \log_{b} b} + \frac{1}{\log_{c} a b + \log_{c} c} \\ = \frac{1}{\log_{a} (a b c)} + \frac{1}{\log_{b} (a b c)} + \frac{1}{\log_{c} (a b c)} \\ = \log_{a b c} a + \log_{a b c} b + \log_{a b c} c \\ = \log_{a b c} (a b c) \\ = 1 \end{aligned}

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