\({3^{{x^2} - 4}}{.5^{x + m}} = 3 \Leftrightarrow {3^{{x^2} - 5}}{.5^{x + m}} = 1 \)

\(\Leftrightarrow \left( {{x^2} - 5} \right)\ln 3 + \left( {x + m} \right)\ln 5 = 0\)

\(\Leftrightarrow {x^2}.\ln 3 + x.\ln 5 - 5\ln 3 + m\ln 5 = 0\,(*)\)

Giải sử (*) có nghiệm \(x_1, x_2\). Áp dụng định lý Vi-et ta có:

\(\left\{ \begin{array}{l} {x_1} + {x_2} = \frac{{\ln 5}}{{\ln 3}}\\ {x_1}.{x_2} = - 5 + m\frac{{\ln 5}}{{\ln 3}} \end{array} \right.\)
Khi đó: 

\(\left| {{x_1} - {x_2}} \right| = {\log _3}5 \)

\(\Leftrightarrow {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = \frac{{{{\ln }^2}5}}{{{{\ln }^2}3}} \)

\(\Leftrightarrow {\left( { - \frac{{\ln 5}}{{\ln 3}}} \right)^2} - 4\left( { - 5 + \frac{{m\ln 5}}{{\ln 3}}} \right)\)\(= \frac{{{{\ln }^2}5}}{{{{\ln }^2}3}}\)
\(\Leftrightarrow \frac{{m\ln 5}}{{\ln 3}} = 5 \Leftrightarrow m = 5{\log _5}3\).