a) Ta có:
\(\eqalign{
& {\log _7}{{a + b} \over 3} = {1 \over 2}(log_7a + \log _7b) \cr
& \Leftrightarrow 2lo{g_7}{{a + b} \over 3} = {\log _7}(ab) \cr
& \Leftrightarrow {({{a + b} \over 3})^2} = ab \cr
& \Leftrightarrow {a^2} + 2ab + {b^2} = 9ab \Leftrightarrow {a^2} + {b^2} = 7ab\,\,(đpcm) \cr} \)
b) Ta có:
\(\eqalign{
& {\log _{a\sqrt b }}{{\root 3 \of a } \over {\sqrt {{b^3}} }} = {{{{\log }_a}{{\root 3 \of a } \over {\sqrt {{b^3}} }}} \over {{{\log }_a}a\sqrt b }} = {{{{\log }_a}\root 3 \of a - {{\log }_a}\sqrt {{b^3}} } \over {{{\log }_a}a + {{\log }_a}\sqrt b }} \cr
& = {{{1 \over 3} - {3 \over 2}{{\log }_a}b} \over {1 + {1 \over 2}{{\log }_a}b}} = {{{1 \over 3} - {3 \over 2}\sqrt 3 } \over {1 + {1 \over 2}\sqrt 3 }} \cr
& = {{2 - 9\sqrt 3 } \over {6 + 3\sqrt 3 }} \cr} \)
