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$log_{ \frac{1}{3} }(x- \frac{1}{6} ) =1- log_{ \frac{1}{3} } (x+ \frac{1}{2} )$
ОДЗ:
$\left \{ {{x- \frac{1}{6}\ \textgreater \ 0 } \atop {x+ \frac{1}{2}\ \textgreater \ 0 }} \right. , \left \{ {{x\ \textgreater \ \frac{1}{6} } \atop {x\ \textgreater \ - \frac{1}{2} }} \right. =\ \textgreater \ x\ \textgreater \ \frac{1}{6}$
$log_{ \frac{1}{3} } (x- \frac{1}{6} )+ log_{ \frac{1}{3} }(x+ \frac{1}{2} ) =1$
$log_{ \frac{1}{3} } ((x- \frac{1}{6} )*(x+ \frac{1}{2} ))=1$
$(x- \frac{1}{6} )*(x+ \frac{1}{2} )=( \frac{1}{3} ) ^{1}$
$x^{2} - \frac{1}{6} x+ \frac{1}{2} x- \frac{1}{12}= \frac{1}{3} |*12$
12x²+4x-5=0
x₁=-5/6 посторонний корень
x₂=1/2
ответ: х=0,5