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If $x=\log _a b c, y=\log _b c a, z=\log _c a b$, then the value of $\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}$ will be
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Hints: $1+x=\log _a a+\log _a b c=\log _a a b c$
$\begin{aligned}
& \frac{1}{1+x}=\log _{a b c} a, \text { Similarly } \frac{1}{1+y}=\log _{a b c} b \\
& \frac{1}{1+z}=\log _{a b c} c \text {, Ans. }=\log _{(a b c)} a b c=1
\end{aligned}$
$\begin{aligned}
& \frac{1}{1+x}=\log _{a b c} a, \text { Similarly } \frac{1}{1+y}=\log _{a b c} b \\
& \frac{1}{1+z}=\log _{a b c} c \text {, Ans. }=\log _{(a b c)} a b c=1
\end{aligned}$
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