Cho các số thực dương a, b, c thỏa mãn a^log7=27

Phát biểu	Đúng	Sai
\(\sqrt[3]{{{a^{{{\left( {{{\log }_3}7} \right)}^2}}}}} = 14\)	¡	¤
\({c^{{{\left( {{{\log }_{11}}25} \right)}^2}}} = 5\)	¤	¡
\(\sqrt[3]{{{a^{{{\left( {{{\log }_3}7} \right)}^2}}}}} + \sqrt {{b^{{{\left( {{{\log }_7}11} \right)}^2}}}}  + {c^{{{\left( {{{\log }_{11}}25} \right)}^2}}} = 23\)	¤	¡

Giải thích

\(\sqrt[3]{{{a^{{{\left( {{{\log }_3}7} \right)}^2}}}}} = \sqrt[3]{{{{\left( {{a^{{{\log }_3}7}}} \right)}^{{{\log }_3}7}}}} = \sqrt[3]{{{{27}^{{{\log }_3}7}}}} = \sqrt[3]{{{{\left( {{3^{{{\log }_3}7}}} \right)}^3}}} = 7\)

\(\sqrt {{b^{{{\left( {{{\log }_7}11} \right)}^2}}}}  = \sqrt {{{\left( {{b^{{{\log }_7}11}}} \right)}^{{{\log }_7}11}}}  = \sqrt {{{49}^{{{\log }_7}11}}}  = \sqrt {{{\left( {{7^{{{\log }_7}11}}} \right)}^2}}  = 11.\)

\({c^{{{\left( {{{\log }_{11}}25} \right)}^2}}} = {\left( {{c^{{{\log }_{11}}25}}} \right)^{{{\log }_{11}}25}} = {(\sqrt {11} )^{{{\log }_{11}}25}} = \sqrt {{{11}^{{{\log }_{11}}25}}}  = \sqrt {25}  = 5\) 

Vậy \(\sqrt[3]{{{a^{{{\left( {{{\log }_3}7} \right)}^2}}}}} + \sqrt {{b^{{{\left( {{{\log }_7}11} \right)}^2}}}}  + {c^{{{\left( {{{\log }_{11}}25} \right)}^2}}} = 7 + 11 + 5 = 23\).