Với giá trị nào của $k$ thì phương trình ${9^x} -\left( {k - 1} \right){3^x} + 2k = 0  \,\,\,\left( 1 \right)$ có $1$ nghiệm duy  nhất
$\left( 1 \right) \Leftrightarrow {3^{2x}} - \left( {k - 1} \right){3^x} + 2k = 0$
Đặt $t = {3^x},\,t > 0$. Ta có:
    ${t^2} - \left( {k - 1} \right)t + 2k = 0    \left( 2 \right)$
($1$)    có nghiệm duy nhất $ \Leftrightarrow $($2$) có $1$ nghiệm dương duy nhất
$\begin{array}{l}
 \Leftrightarrow \left[ \begin{array}{l}
{t_1} = {t_2} > 0\\
{t_1} < 0 < {t_2}
\end{array} \right.\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\Delta  = 0\\
S > 0
\end{array} \right.\\
P < 0
\end{array} \right.\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{\left( {k - 1} \right)^2} - 8k = 0\\
k - 1 > 0
\end{array} \right.\\
2k < 0
\end{array} \right. \\
 \Leftrightarrow \left[ \begin{array}{l}
k = 5 + 2\sqrt 6 \,; \left( {k = 5 - 2\sqrt 6  < 1\,\,\,\,loạ i} \right)\\
k < 0
\end{array} \right.\Leftrightarrow \left[ \begin{array}{l}
k = 5 + 2\sqrt 6 \\
k < 0
\end{array} \right.
\end{array}$

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