Lời giải:.png)
Ta có:
\(\begin{array}{*{20}{l}}{\cos B = \frac{{A{B^2} + B{C^2} - A{C^2}}}{{2AB.BC}}}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = \frac{{{4^2} + {6^2} - {{\left( {2\sqrt 7 } \right)}^2}}}{{2.4.6}} = \frac{1}{2}}\end{array}\)
Vì MC = 2MB, BC = 6 nên \(BM = \frac{1}{3}BC = \frac{1}{3}.6 = 2.\)
Áp dụng định lí cosin trong tam giác ABM ta có:
\(\begin{array}{*{20}{l}}{A{M^2} = A{B^2} + B{M^2} - 2AB.BM.\cos B}\\{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = {4^2} + {2^2} - 2.4.2.\frac{1}{2} = 12}\\{ \Rightarrow AM = 2\sqrt 3 .}\end{array}\)
Chọn C.
