Đáp án đúng: A
Giải chi tiết:Ta có:
\(\eqalign{ & \,\,\,\,\,m_a^2 = {{{b^2} + {c^2}} \over 2} - {{{a^2}} \over 4} \cr & \,\,\,\,\,m_b^2 = {{{a^2} + {c^2}} \over 2} - {{{b^2}} \over 4} \cr & \,\,\,\,\,m_c^2 = {{{a^2} + {b^2}} \over 2} - {{{c^2}} \over 4} \cr & \Rightarrow m_a^2 + m_b^2 + m_c^2 = {{{b^2} + {c^2}} \over 2} + {{{a^2} + {c^2}} \over 2} + {{{a^2} + {b^2}} \over 2} - {{{a^2}} \over 4} - {{{b^2}} \over 4} - {{{c^2}} \over 4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{2\left( {{a^2} + {b^2} + {c^2}} \right)} \over 2} - {{{a^2} + {b^2} + {c^2}} \over 4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {3 \over 4}\left( {{a^2} + {b^2} + {c^2}} \right) = {{75} \over 2} \cr} \)
Chọn A
