Cho hệ phương trình :
               $\left\{ \begin{array}{l}
9{x^2} - 4{y^2} = 5\\
{\log _m}\left( {3x + 2y} \right) - {\log _3}\left( {3x - 2y} \right) = 1
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$
$1)$  Giải ($1$) khi $m = 5$
$2)$  Tìm giá trị lớn nhất của tham số $m$ sao cho hệ ($1$) có nghiệm $\left( {x,\,y} \right)$ thỏa mãn :
      $3x + 2y \le 5$
$1$) Đặt $\begin{array}{l}
X = 3x + 2y\\
Y = 3x - 2y
\end{array}$   ,Điều kiện:   $\left\{ \begin{array}{l}
X,Y > 0\\
1 \ne m > 0
\end{array} \right.$
Hệ ($1$) $ \Leftrightarrow \left\{ \begin{array}{l}
XY = 5\\
{\log _m}X - {\log _3}Y = 1\,\,
\end{array} \right.\,\,\, \Leftrightarrow \,\left\{
\begin{array}{l}
Y = \frac{5}{X}\\
{\log _m}X - {\log _3}\frac{5}{X} = 1
\end{array} \right.$
$\begin{array}{l}
 \Leftrightarrow \,\,\left\{ \begin{array}{l}
Y = \frac{5}{X}\\
{\log _m}X - {\log _3}5 + {\log _3}X = 1
\end{array} \right.\\
 \Leftrightarrow \,\,\left\{ \begin{array}{l}
Y = \frac{5}{X}\\
\frac{{{{\log }_3}X}}{{{{\log }_3}m}} + {\log _3}X = {\log _3}5 + 1
\end{array} \right.\\
 \Leftrightarrow \,\,\left\{ \begin{array}{l}
Y = \frac{5}{X}\\
{\log _3}X = \frac{{\left( {{{\log }_3}5 + 1} \right){{\log }_3}m}}{{1 + {{\log }_3}m}}
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,(2)\,\,\,\,\,\,\,\,\,\,\,\,m \ne \frac{1}{3}
\end{array}$
($m = \frac{1}{3}$hệ vô nghiệm)
Với $m = 5$  ta có : $\left\{ \begin{array}{l}
Y = \frac{5}{X}\\
{\log _3}X = {\log _3}5
\end{array} \right. \Leftrightarrow \,\,\,\,\left\{ \begin{array}{l}
X = 5\\
Y = 1
\end{array} \right.$
$ \Leftrightarrow \left\{ \begin{array}{l}
3x + 2y = 5\\
3x - 2y = 1
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y = 1
\end{array} \right.$
Vậy với $m = 5$ hệ có nghiệm là $\left( {1,\,1} \right)$
$2)$ $3x + 2y \le 5\,\,\,\, \Leftrightarrow X \le 5\,\,\, \Leftrightarrow \,\,\,{\log _3}X \le {\log _3}5$
$ \Leftrightarrow \frac{{\left( {{{\log }_3}5 + 1} \right){{\log }_3}m}}{{1 + {{\log }_3}m}} \le
{\log _3}5\,\,\,\, \Leftrightarrow \frac{{{{\log }_3}m - {{\log }_3}5}}{{1 + {{\log }_3}m}}\,\,\,
\le 0\,\,\,\,\,\,\,(3)$
Đặt $M = {\log _3}m$ ta có $(3) \Leftrightarrow \,\,\,\,\,\,\, - 1 < M < {\log _3}5$
$ \Leftrightarrow \,\,\,\, - 1 < {\log _3}m < {\log _3}5\,\,\,\, \Leftrightarrow \,\,\,\frac{1}{3} < m
\le 5$
   Vậy giá trị lớn nhất của m là $5$ thì $(1)$ có nghiệm $\left( {x,\,y} \right)$ thỏa mãn :            
$3x + 2y \le 5$
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