Chứng minh rằng:
a) \({\rm{lo}}{{\rm{g}}_a}\left( {x + \sqrt {{x^2} - 1} } \right) + {\rm{lo}}{{\rm{g}}_a}\left( {x - \sqrt {{x^2} - 1} } \right) = 0\);
b) \({\rm{ln}}\left( {1 + {e^{2x}}} \right) = 2x + {\rm{ln}}\left( {1 + {e^{ - 2x}}} \right)\).
Áp dụng quy tắc tính logarit
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\({\log _a}(MN) = {\log _a}M + {\log _a}N;\)
Biến đổi \(1 + {e^{2x}}{e^{2x}} = \left( {1 + {e^{ - 2x}}} \right)\)
a) \({\rm{lo}}{{\rm{g}}_a}\left( {x + \sqrt {{x^2} - 1} } \right) + {\rm{lo}}{{\rm{g}}_a}\left( {x - \sqrt {{x^2} - 1} } \right) = {\rm{lo}}{{\rm{g}}_a}\left[ {\left( {x + \sqrt {{x^2} - 1} } \right)\left( {x - \sqrt {{x^2} - 1} } \right)} \right]\)
\({\rm{ = lo}}{{\rm{g}}_a}\left( {{x^2} - \left( {{x^2} - 1} \right)} \right) = \)\( = {\rm{lo}}{{\rm{g}}_a}1 = 0\).
b) \({\rm{ln}}\left( {1 + {e^{2x}}} \right) = {\rm{ln}}\left[ {{e^{2x}}\left( {1 + {e^{ - 2x}}} \right)} \right] = {\rm{ln}}{e^{2x}} + {\rm{ln}}\left( {1 + {e^{ - 2x}}} \right)\)\( = 2x + {\rm{ln}}\left( {1 + {e^{ - 2x}}} \right){\rm{.\;}}\)