Đáp án:D
Giải thích các bước giải:
$\begin{array}{l}
3{\log _{27}}\left( {2{x^2} - \left( {m + 3} \right)x + 1 - m} \right) + {\log _{\frac{1}{3}}}\left( {{x^2} - x + 1 - 3m} \right) = 0\\
Đkxđ:\left\{ \begin{array}{l}
2{x^2} - \left( {m + 3} \right)x + 1 - m > 0\\
{x^2} - x + 1 - 3m > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{\Delta _1} < 0\\
{\Delta _2} < 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {m + 3} \right)^2} - 8\left( {1 - m} \right) < 0\\
1 - 4\left( {1 - 3m} \right) < 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{m^2} + 14m + 1 < 0\\
12m - 3 < 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- 13,07 < m < - 0,07\\
m < 0,25
\end{array} \right.\\
\Rightarrow - 13,07 < m < 0,25\left( 1 \right)\\
Pt \Leftrightarrow {\log _3}\left( {2{x^2} - \left( {m + 3} \right)x + 1 - m} \right) = {\log _3}\left( {{x^2} - x + 1 - 3m} \right)\\
\Leftrightarrow 2{x^2} - \left( {m + 3} \right)x + 1 - m = {x^2} - x + 1 - 3m\\
\Leftrightarrow {x^2} - \left( {m + 2} \right)x + 2m = 0\\
\Rightarrow \left\{ \begin{array}{l}
\Delta > 0\\
{x_1} + {x_2} = m + 2\\
{x_1}.{x_2} = 2m
\end{array} \right. \Rightarrow \left\{ {{{\left( {m + 2} \right)}^2} - 8m > 0 \Rightarrow m \ne 2\left( 2 \right)} \right.\\
Khi\left| {{x_1} - {x_2}} \right| < 15\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}.{x_2} < {15^2}\\
\Rightarrow {\left( {m + 2} \right)^2} - 4.2m < 225\\
\Rightarrow {m^2} - 4m - 221 < 0\\
\Rightarrow - 13 < m < 17\left( 3 \right)\\
Từu\,\left( 1 \right);\left( 2 \right)\left( 3 \right) \Rightarrow - 13 < m < 0,25
\end{array}$
