Giải thích các bước giải:
$\begin{array}{l}
a)\overrightarrow {AB} = \overrightarrow {AK} + \overrightarrow {KM} + \overrightarrow {MB} \\
\Leftrightarrow \overrightarrow {AB} = \overrightarrow {AK} + \dfrac{1}{2}\overrightarrow {BA} - \overrightarrow {BM} \\
\Leftrightarrow \overrightarrow {AB} = \overrightarrow {AK} - \dfrac{1}{2}\overrightarrow {AB} - \overrightarrow {BM} \\
\Leftrightarrow \dfrac{3}{2}\overrightarrow {AB} = \overrightarrow {AK} - \overrightarrow {BM} \\
\Leftrightarrow \overrightarrow {AB} = \dfrac{2}{3}\overrightarrow {AK} - \dfrac{2}{3}\overrightarrow {BM} \\
b)\overrightarrow {BC} = 2\overrightarrow {BK} \\
\Leftrightarrow \overrightarrow {BC} = 2\left( {\overrightarrow {BA} + \overrightarrow {AK} } \right)\\
\Leftrightarrow \overrightarrow {BC} = 2\left( { - \overrightarrow {AB} + \overrightarrow {AK} } \right)\\
\Leftrightarrow \overrightarrow {BC} = 2\left( { - \dfrac{2}{3}\overrightarrow {AK} + \dfrac{2}{3}\overrightarrow {BM} + \overrightarrow {AK} } \right)\\
\Leftrightarrow \overrightarrow {BC} = 2\left( {\dfrac{1}{3}\overrightarrow {AK} + \dfrac{2}{3}\overrightarrow {BM} } \right)\\
\Leftrightarrow \overrightarrow {BC} = \dfrac{2}{3}\overrightarrow {AK} + \dfrac{4}{3}\overrightarrow {BM} \\
c)\overrightarrow {CA} = \overrightarrow {CB} + \overrightarrow {BA} \\
\Leftrightarrow \overrightarrow {CA} = - \overrightarrow {BC} - \overrightarrow {AB} \\
\Leftrightarrow \overrightarrow {CA} = - \left( {\dfrac{2}{3}\overrightarrow {AK} + \dfrac{4}{3}\overrightarrow {BM} } \right) - \left( {\dfrac{2}{3}\overrightarrow {AK} - \dfrac{2}{3}\overrightarrow {BM} } \right)\\
\Leftrightarrow \overrightarrow {CA} = - \dfrac{4}{3}\overrightarrow {AK} - \dfrac{2}{3}\overrightarrow {BM}
\end{array}$
