Đặt \(\overrightarrow {KA} = x\overrightarrow {AB} \)

Ta có: \(\frac = \frac = \frac{4}{{1,5}} = \frac{8}{3}\)⇒ \(\frac = \frac{8}\)

\[\overrightarrow {AM} = \overrightarrow {AB} + \overrightarrow {BM} = \overrightarrow {AB} + \frac{1}{2}\overrightarrow {BC} \]

⇒ \[\overrightarrow {AI} = \frac{8}\overrightarrow {AM} = \frac{8}\overrightarrow {AB} + \frac{4}\overrightarrow {BC} \]

\(\overrightarrow {BI} = \overrightarrow {BA} + \overrightarrow {AI} = \frac{{ - 3}}\overrightarrow {AB} + \frac{4}\overrightarrow {BC} \)

\(\overrightarrow {KM} = \overrightarrow {KA} + \overrightarrow {AM} = x\overrightarrow {AB} + \overrightarrow {AB} + \frac{1}{2}\overrightarrow {BC} = \left( {1 + x} \right)\overrightarrow {AB} + \frac{1}{2}\overrightarrow {BC} \)

Vì KM vuông góc với BI nên \(\overrightarrow {BI} .\overrightarrow {KM} = 0\)

Hay \(\left( {\frac{{ - 3}}\overrightarrow {AB} + \frac{4}\overrightarrow {BC} } \right)\left[ {\left( {1 + x} \right)\overrightarrow {AB} + \frac{1}{2}\overrightarrow {BC} } \right] = 0\)

Đặt \(\overrightarrow {AB} = \overrightarrow a ,\overrightarrow {CB} = \overrightarrow b \)

Ta có: \(\frac{{ - 3\left( {1 + x} \right)}}{\overrightarrow a ^2} + \frac{3}\overrightarrow a .\overrightarrow b - \frac{{4\left( {1 + x} \right)}}\overrightarrow a .\overrightarrow b + \frac{2}{\overrightarrow b ^2} = 0\)

⇔ \(\frac{{ - 3\left( {1 + x} \right)}}{.2^2} + \left( {\frac{{ - 4\left( {1 + x} \right)}} + \frac{3}} \right).\overrightarrow a .\overrightarrow b + \frac{2}{.3^2} = 0\)

Lại có: \(\overrightarrow a .\overrightarrow b = 2.3.\frac{{{2^2} + {3^2} - {4^2}}} = \frac{{ - 3}}{2}\)

⇒ \(\frac{{ - 12 - 12x}} + \left( {\frac} \right).\frac{{ - 3}}{2} + \frac = 0\)

⇔ x = \(\frac{8}\)

Vậy \(\frac\)= \(\frac{8}\).