Đáp án:
\(5 < m < \frac{{21}}{4}\)
Giải thích các bước giải:
\[\begin{array}{l}
{\log _3}\left( {1 - {x^2}} \right) + {\log _{\frac{1}{3}}}\left( {x + m - 4} \right) = 0\,\,\,\,\,\left( * \right)\,\,\,\,\left( {DK:\,\,\, - \,1 < x < 1,\,\,\,x + m - 4 > 0} \right)\\
\Leftrightarrow {\log _3}\left( {1 - {x^2}} \right) - {\log _3}\left( {x + m - 4} \right) = 0\,\,\,\\
\Leftrightarrow \frac{{1 - {x^2}}}{{x + m - 4}} = {3^0} = 1\\
\Leftrightarrow 1 - {x^2} = x + m - 4\\
\Leftrightarrow {x^2} + x + m - 5 = 0\,\,\,\,\,\left( 1 \right)\\
pt\,\,\left( * \right)\,\,\,co\,\,\,2\,\,\,nghiem\,\,\,pb \Leftrightarrow \left( 1 \right)\,\,\,co\,\,\,2\,\,\,nghiem\,\,\,pb\,\,\,\, - 1 < x < 1\\
\Leftrightarrow \left\{ \begin{array}{l}
\Delta > 0\\
- 1 < {x_1} < {x_2} < 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
1 - 4m + 20 > 0\\
{x_1} + {x_2} > - 2\\
\left( {{x_1} + 1} \right)\left( {{x_2} + 1} \right) > 0\\
{x_1} + {x_2} < 2\\
\left( {{x_1} - 1} \right)\left( {{x_2} - 1} \right) > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
4m < 21\\
- 1 > - 2\,\,\,\left( {luon\,\,dung} \right)\\
{x_1}{x_2} + {x_1} + {x_2} + 1 > 0\\
- 1 > 2\,\,\,\left( {luon\,\,\,dung} \right)\\
{x_1}{x_2} - \left( {{x_1} + {x_2}} \right) + 1 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m < \frac{{21}}{4}\\
m - 5 - 1 + 1 > 0\\
m - 5 + 1 + 1 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m < \frac{{21}}{4}\\
m > 5\\
m > 3
\end{array} \right. \Leftrightarrow 5 < m < \frac{{21}}{4}.
\end{array}\]
