giải toán 10: cho tam giác ABC có vectoMB=3vectoMC , vecto BE =2 vecto EC và vecto AN = 1\3 vecto AB . biết AE giao MN =I a) tính AI \ IE

$\begin{array}{l} \overrightarrow {AE} = \overrightarrow {AB} + \overrightarrow {BE} = \overrightarrow {AB} + \frac{2}{3}\overrightarrow {BC} = \overrightarrow {AB} + \frac{2}{3}.\frac{2}{3}\overrightarrow {BM} = \overrightarrow {AB} + \frac{4}{9}\overrightarrow {BM} \\ \overrightarrow {AI} = \overrightarrow {AN} + \overrightarrow {NI} = \frac{1}{3}\overrightarrow {AB} + k\overrightarrow {NM} = \frac{1}{3}\overrightarrow {AB} + k\left( {\overrightarrow {NB} + \overrightarrow {BM} } \right) = \frac{1}{3}\overrightarrow {AB} + k.\frac{2}{3}\overrightarrow {AB} + k\overrightarrow {BM} = \left( {\frac{{1 + 2k}}{3}} \right)\overrightarrow {AB} + k\overrightarrow {BM} \\ A,I,E\,thang\,hang \Leftrightarrow \overrightarrow {AI} = l\overrightarrow {AE} \Leftrightarrow \left\{ \begin{array}{l} \frac{{1 + 2k}}{3} = l\\ k = \frac{4}{9}l \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} l = \frac{9}{{19}}\\ k = \frac{4}{{19}} \end{array} \right.\\ \Rightarrow \overrightarrow {AI} = \frac{9}{{19}}\overrightarrow {AE} = \frac{9}{{19}}\left( {\overrightarrow {AI} + \overrightarrow {IE} } \right) \Rightarrow \frac{{10}}{{19}}\overrightarrow {AI} = \frac{9}{{19}}\overrightarrow {IE} \Rightarrow \frac{{AI}}{{IE}} = \frac{9}{{10}} \end{array}$