tìm m để pt sau có 2 nghiệm phân biệt log3(1-x^2) +log1/3(x+m-4)=0

Đáp án:

$5 < m < \frac{{21}}{4}$

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