Ta có:

a) 

$\overrightarrow{AP} = \overrightarrow{AD} + \overrightarrow{DP} = \overrightarrow{AD} + \frac{2}{3}\overrightarrow{DM} = \overrightarrow{AD} + \frac{2}{3}(\overrightarrow{AM} - \overrightarrow{AD}) = \frac{1}{3}\overrightarrow{AD} + \frac{2}{3}\overrightarrow{AM} = \frac{1}{3}\overrightarrow{AD} + \frac{2}{3} . \frac{1}{2}\overrightarrow{AB} = \frac{1}{3}\overrightarrow{u} + \frac{1}{3}\overrightarrow{v}$

Có BQ = xQN

=>$ \overrightarrow{BQ} = x\overrightarrow{QN}$

<=> $\overrightarrow{AQ} - \overrightarrow{AB} = x(\overrightarrow{AN} - \overrightarrow{AQ})$

<=> $(x + 1)\overrightarrow{AQ} = \overrightarrow{AB} x\overrightarrow{AN}$

$<=> (x + 1)\overrightarrow{AQ} = \overrightarrow{AB} + (\overrightarrow{AD} - \overrightarrow{DN}) = x\overrightarrow{AD} + \overrightarrow{AB} + x.\frac{1}{2}\overrightarrow{AB}$

$<=> (x + 1)\overrightarrow{AQ} = x\overrightarrow{AD} + (\frac{1}{2}x + 1)\overrightarrow{AB}$

$<=> (x + 1)\overrightarrow{AQ} = x\overrightarrow{u} + (\frac{1}{2}x + 1)\overrightarrow{v}$

$<=> \overrightarrow{AQ} = \frac{x + 2}{2(x + 1)}\overrightarrow{u} + \frac{x}{x + 1}\overrightarrow{v}$

b) 

A, P, Q thằng hàng nên:

$<=> \overrightarrow{AP} và  \overrightarrow{AQ}$ cùng phương

$<=> \frac{x + 2}{2(x + 1)} : \frac{2}{3} = \frac{x}{x + 1} : \frac{2}{3}$ với x $\neq$ -1

$<=> \frac{x + 2}{2} = x$

$<=> x = 2$ thỏa mãn điều kiện