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

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1. Đáp án:

   \(5 < m < \frac{{21}}{4}\)

   Giải thích các bước giải:

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

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