Đáp án đúng: A
Giải chi tiết:Ta có: \({{{b^3} + {c^3} - {a^3}} \over {b + c - a}} = {a^2}\)
\(\eqalign{ & \Leftrightarrow {b^3} + {c^3} - {a^3} = {a^2}\left( {b + c - a} \right) \cr & \Leftrightarrow {b^3} + {c^3} - {a^3} = {a^2}b + {a^2}c - {a^3} \cr & \Leftrightarrow {b^3} + {c^3} - {a^2}b - {a^2}c = 0 \cr & \Leftrightarrow \left( {b + c} \right)\left( {{b^2} - bc + {c^2}} \right) - {a^2}\left( {b + c} \right) = 0 \cr & \Leftrightarrow \left( {b + c} \right)\left( {{b^2} - bc + {c^2} - {a^2}} \right) = 0 \cr & \Leftrightarrow {b^2} + {c^2} - {a^2} - bc = 0\,\,\,\,(b + c > 0)\,\,\left( * \right) \cr} \)
Áp dụng định lý cosin cho \(\Delta ABC\) ta có:
\(\eqalign{ & \,\,\,\,\,\,\,\,{a^2} = {b^2} + {c^2} - 2bc\cos A \cr & \Leftrightarrow {b^2} + {c^2} - {a^2} = 2bc\cos \,A \cr & \left( * \right) \Leftrightarrow 2bc\cos A - bc = 0 \cr & \,\,\,\,\,\,\, \Leftrightarrow 2bc\cos A = bc \cr & \,\,\,\,\,\,\, \Leftrightarrow \cos A = {1 \over 2} \cr & \,\,\,\,\,\,\, \Rightarrow \widehat A = {60^0}(dpcm) \cr} \)
Chọn A
