Cho \(\left( P \right):y = {x^2} + bx + 1\) đi qua điểm \(A\left( { - 1;3} \right)\). Khi đó

a) \(\overrightarrow {MD}  = \overrightarrow {MG}  + \overrightarrow {GD} \).

b) Ta có: \(\overrightarrow {AG}  = \frac{2}{3}\overrightarrow {AM}  = \frac{2}{3} \cdot \frac{1}{2}(\overrightarrow {AB}  + \overrightarrow {AC} ) = \frac{1}{3}\overrightarrow {AB}  + \frac{1}{3}\overrightarrow {AC} \).

c) Ta có: \(\overrightarrow {CD}  = \overrightarrow {CB}  + \overrightarrow {BD}  = \overrightarrow {AB}  - \overrightarrow {AC}  + \frac{4}{3}\overrightarrow {BN} \).

d) Ta có: \(\overrightarrow {MD}  = \overrightarrow {MG}  + \overrightarrow {GD}  =  - \frac{1}{3}\overrightarrow {AM}  + \frac{2}{3}\overrightarrow {BN}  =  - \frac{1}{3} \cdot \frac{1}{2}(\overrightarrow {AB}  + \overrightarrow {AC} ) + \frac{2}{3}(\overrightarrow {BA}  + \overrightarrow {AN} )\)

\( =  - \frac{1}{6}\overrightarrow {AB}  - \frac{1}{6}\overrightarrow {AC}  - \frac{2}{3}\overrightarrow {AB}  + \frac{2}{3} \cdot \frac{1}{2}\overrightarrow {AC}  =  - \frac{5}{6}\overrightarrow {AB}  + \frac{1}{6}\overrightarrow {AC} {\rm{. }}\)