Đáp án:
Vậy \(1 \le m \le \frac{4}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
\left( {m - 1} \right){.3^{2x}} + 2\left( {m - 3} \right){.3^x} + m + 3 = 0\,\,\left( * \right)\\
Dat\,t = {3^x} > 0\,thi\,\left( * \right)\,tro\,thanh\,:\\
\left( {m - 1} \right).{t^2} + 2\left( {m - 3} \right)t + m + 3 = 0\,\,\left( {**} \right)\\
+ )\,Neu\,m = 1\,thi\,\left( {**} \right)\,la\, - 4t + 4 = 0 \Leftrightarrow t = 1 > 0\,\,\left( {TM} \right)\\
\Rightarrow Phuong\,trinh\,co\,nghiem.\\
+ )\,Neu\,m \ne 1\,thi\,\left( * \right)\,co\,nghiem\, \Leftrightarrow \left( {**} \right)\,co\,it\,nhat\,1\,nghiem\,duong\\
- TH1:\,\left( {**} \right)\,co\,2\,nghiem\,trai\,dau \Leftrightarrow \left( {m - 1} \right)\left( {m + 3} \right) < 0 \Leftrightarrow - 3 < m < 1\\
- TH2:\left( {**} \right)\,co\,2\,nghiem\,duong\, \Leftrightarrow \left\{ \begin{array}{l}
\Delta ' \ge 0\\
S > 0\\
P > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {m - 3} \right)^2} - \left( {m - 1} \right)\left( {m + 3} \right) \ge 0\\
- \frac{{2\left( {m - 3} \right)}}{{m - 1}} > 0\\
\frac{{m + 3}}{{m - 1}} > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
- 8m + 12 \ge 0\\
\frac{{m - 3}}{{m - 1}} < 0\\
\frac{{m + 3}}{{m - 1}} > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \le \frac{4}{3}\\
1 < m < 3\\
\left[ \begin{array}{l}
m > 1\\
m < - 3
\end{array} \right.
\end{array} \right. \Leftrightarrow 1 < m \le \frac{4}{3}\\
Vay\,1 < m \le \frac{4}{3}
\end{array}\)
