Log3(2x+1)-log1/9(x-1)2=1

Đáp án:

\(\left[ \begin{array}{l}
x = \frac{{1 + \sqrt {17} }}{4}\\
x = 0\\
x = \frac{1}{2}
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
{\log _3}(2x + 1) - {\log _{\frac{1}{9}}}{(x - 1)^2} = 0\\
DKXD:x > \frac{{ - 1}}{2}\\
 \Leftrightarrow {\log _3}(2x + 1) + \frac{1}{2}{\log _3}{(x - 1)^2} = 0\\
 \Leftrightarrow {\log _3}(2x + 1) + {\log _3}|x - 1| = 0\\
 \Leftrightarrow (2x + 1)|x - 1| = 1\\
 \Leftrightarrow \left[ \begin{array}{l}
(2x + 1)(x - 1) = 1\\
(2x + 1)(1 - x) = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{1 + \sqrt {17} }}{4}\\
x = 0\\
x = \frac{1}{2}
\end{array} \right.
\end{array}\)