Question

Đã trả lời bởi chuyên gia

Answer 1

Đáp án

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

Giải thích

\(\sqrt[3]{{{a^{{{\left( {{\rm{lo}}{{\rm{g}}_3}7} \right)}^2}}}}} = \sqrt[3]{{{{\left( {{a^{{\rm{lo}}{{\rm{g}}_3}7}}} \right)}^{{\rm{lo}}{{\rm{g}}_3}7}}}} = \sqrt[3]{{{{27}^{{\rm{lo}}{{\rm{g}}_3}7}}}} = \sqrt[3]{{{{\left( {{3^{{\rm{lo}}{{\rm{g}}_3}7}}} \right)}^3}}} = 7\).

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

\({c^{{{\left( {{\rm{lo}}{{\rm{g}}_{11}}25} \right)}^2}}} = {\left( {{c^{{\rm{lo}}{{\rm{g}}_{11}}25}}} \right)^{{\rm{lo}}{{\rm{g}}_{11}}25}} = {(\sqrt {11} )^{{\rm{lo}}{{\rm{g}}_{11}}25}} = \sqrt {{{11}^{{\rm{lo}}{{\rm{g}}_{11}}25}}} = \sqrt {25} = 5\).

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

...Xem thêm

Answer 2

Đáp án

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

Giải thích

\(\sqrt[3]{{{a^{{{\left( {{\rm{lo}}{{\rm{g}}_3}7} \right)}^2}}}}} = \sqrt[3]{{{{\left( {{a^{{\rm{lo}}{{\rm{g}}_3}7}}} \right)}^{{\rm{lo}}{{\rm{g}}_3}7}}}} = \sqrt[3]{{{{27}^{{\rm{lo}}{{\rm{g}}_3}7}}}} = \sqrt[3]{{{{\left( {{3^{{\rm{lo}}{{\rm{g}}_3}7}}} \right)}^3}}} = 7\).

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

\({c^{{{\left( {{\rm{lo}}{{\rm{g}}_{11}}25} \right)}^2}}} = {\left( {{c^{{\rm{lo}}{{\rm{g}}_{11}}25}}} \right)^{{\rm{lo}}{{\rm{g}}_{11}}25}} = {(\sqrt {11} )^{{\rm{lo}}{{\rm{g}}_{11}}25}} = \sqrt {{{11}^{{\rm{lo}}{{\rm{g}}_{11}}25}}} = \sqrt {25} = 5\).

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

2 câu trả lời 14

Câu hỏi hot cùng chủ đề

Bài viết mới nhất Lớp 12