Đáp án:
\[\left[ \begin{array}{l}
m = 25\\
m = \frac{1}{{25}}
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{25^x} - \left( {m + 1} \right){5^x} + m = 0\\
\Leftrightarrow {\left( {{5^x}} \right)^2} - \left( {m + 1} \right){.5^x} + m = 0\\
\Leftrightarrow \left[ {{{\left( {{5^x}} \right)}^2} - {5^x}} \right] - \left( {m{{.5}^x} - m} \right) = 0\\
\Leftrightarrow {5^x}\left( {{5^x} - 1} \right) - m\left( {{5^x} - 1} \right) = 0\\
\Leftrightarrow \left( {{5^x} - 1} \right)\left( {{5^x} - m} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{5^{{x_1}}} = 1\\
{5^{{x_2}}} = m\,\,\,\left( {0 < m \ne 1} \right)
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
{x_1} = 0\\
{5^{{x_2}}} = m
\end{array} \right.\\
{x_1}^2 + {x_2}^2 = 4 \Rightarrow {x_2} = \pm 2 \Rightarrow \left[ \begin{array}{l}
m = 25\\
m = \frac{1}{{25}}
\end{array} \right.\left( {t/m} \right)
\end{array}\)
