Lời giải:

Theo đề ta có: $\overrightarrow{BM}=2\overrightarrow{MC}=-2\overrightarrow{CM}$

$\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}(1)$

$=\overrightarrow{AB}-2\overrightarrow{CM}$

$\overrightarrow{AM}=\overrightarrow{AC}+\overrightarrow{CM}$

$\Rightarrow 2\overrightarrow{AM}=2\overrightarrow{AC}+2\overrightarrow{CM}(2)$

Lấy $(1)+(2)\Rightarrow 3\overrightarrow{AM}=\overrightarrow{AB}+2\overrightarrow{AC}$

$\Rightarrow \overrightarrow{AM}=\frac{1}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}$

Hình vẽ:

Ta có: \(\overrightarrow{MB}=3\overrightarrow{MC}\Rightarrow\overrightarrow{MB}=3\left(\overrightarrow{MB}+\overrightarrow{BC}\right)\)

\(\Rightarrow\overrightarrow{MB}=3\overrightarrow{MB}+3\overrightarrow{BC}\)

\(\Rightarrow-\overrightarrow{MB}=3\overrightarrow{BC}\)

\(\Rightarrow\overrightarrow{BM}=\dfrac{2}{3}\overrightarrow{BC}\). Mà \(\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}\) nên \(\overrightarrow{BM}=\dfrac{2}{3}\left(\overrightarrow{AC}-\overrightarrow{AB}\right)\)

Theo quy tắc 3 điểm, ta có

\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\Rightarrow\overrightarrow{AM}=\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{AC}-\dfrac{3}{2}\overrightarrow{AB}\)

\(\Rightarrow\overrightarrow{AM}=-\dfrac{1}{2}\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{AC}\) hay \(\overrightarrow{AM}=-\dfrac{1}{2}\overrightarrow{u}+\dfrac{3}{2}\overrightarrow{v}\)

Trước hết ta có

= 3 => = 3 ( +)

=> = 3 + 3

=> - = 3

=> =

mà = - nên = (- )

Theo quy tắc 3 điểm, ta có

= + => = + -

=> = - + hay = - +

Tham khảo:

a)  M thuộc cạnh BC nên vectơ \(\overrightarrow {MB} \) và \(\overrightarrow {MC} \) ngược hướng với nhau.

Lại có: MB = 3 MC \( \Rightarrow \overrightarrow {MB}  =  - 3.\overrightarrow {MC} \)

b) Ta có: \(\overrightarrow {AM}  = \overrightarrow {AB}  + \overrightarrow {BM} \)

Mà \(BM = \dfrac{3}{4}BC\) nên \(\overrightarrow {BM}  = \dfrac{3}{4}\overrightarrow {BC} \)

\( \Rightarrow \overrightarrow {AM}  = \overrightarrow {AB}  + \dfrac{3}{4}\overrightarrow {BC} \)

Lại có: \(\overrightarrow {BC}  = \overrightarrow {AC}  - \overrightarrow {AB} \) (quy tắc hiệu)

\( \Rightarrow \overrightarrow {AM}  = \overrightarrow {AB}  + \dfrac{3}{4}\left( {\overrightarrow {AC}  - \overrightarrow {AB} } \right) = \dfrac{1}{4}.\overrightarrow {AB}  + \dfrac{3}{4}.\overrightarrow {AC} \)

Vậy \(\overrightarrow {AM}  = \dfrac{1}{4}.\overrightarrow {AB}  + \dfrac{3}{4}.\overrightarrow {AC} \)

Ta có: \(BC = \frac{{AB}}{{\cos {{30}^o}}} = 3:\frac{{\sqrt 3 }}{2} = 2\sqrt 3 \); \(AC = BC.\sin \widehat {ABC} = 2\sqrt 3 .\sin {30^o} = \sqrt 3 .\)

\(\overrightarrow {BA} .\overrightarrow {BC}  = \left| {\overrightarrow {BA} } \right|.\left| {\overrightarrow {BC} } \right|\cos (\overrightarrow {BA} ,\overrightarrow {BC} ) = 3.2\sqrt 3 .\cos \widehat {ABC} = 6\sqrt 3 .\cos {30^o} = 6\sqrt 3 .\frac{{\sqrt 3 }}{2} = 9.\)

\(\overrightarrow {CA} .\overrightarrow {CB}  = \left| {\overrightarrow {CA} } \right|.\left| {\overrightarrow {CB} } \right|\cos (\overrightarrow {CA} ,\overrightarrow {CB} ) = \sqrt 3 .2\sqrt 3 .\cos \widehat {ACB} = 6.\cos {60^o} = 6.\frac{1}{2} = 3.\)

Lời giải:

\(\overrightarrow{BA}+\overrightarrow{BC}=\overrightarrow{BO}+\overrightarrow{OA}+\overrightarrow{BO}+\overrightarrow{OC}=2\overrightarrow{BO}+(\overrightarrow{OA}+\overrightarrow{OC})\)

\(=2\overrightarrow{BO}\) (do $\overrightarrow{OA}, \overrightarrow{OC}$ là 2 vecto đối)

Và:

\(\overrightarrow{BE}+\overrightarrow{BF}=\overrightarrow{BO}+\overrightarrow{OE}+\overrightarrow{BO}+\overrightarrow{OF}=2\overrightarrow{BO}+(\overrightarrow{OE}+\overrightarrow{OF})\)

\(=2\overrightarrow{BO}\) (do $\overrightarrow{OE}, \overrightarrow{OF}$ là 2 vecto đối)

Vậy \(\overrightarrow{BA}+\overrightarrow{BC}=\overrightarrow{BE}+\overrightarrow{BF}\)