Lời giải

Ta có \(\overrightarrow {A{\rm{E}}} = \overrightarrow {EB} + 3\overrightarrow {EC} \)

\( \Leftrightarrow \overrightarrow {EB} + 3\overrightarrow {EC} - \overrightarrow {A{\rm{E}}} = \overrightarrow 0 \)

\( \Leftrightarrow \overrightarrow {DB} - \overrightarrow {DE} + 3\overrightarrow {DC} - 3\overrightarrow {DE} - \overrightarrow {AD} - \overrightarrow {DE} = \overrightarrow 0 \)

\( \Leftrightarrow \overrightarrow {DB} + 3\overrightarrow {DC} - \overrightarrow {AD} - 5\overrightarrow {DE} = \overrightarrow 0 \)

\( \Leftrightarrow \overrightarrow {DB} + 3\overrightarrow {DC} - \overrightarrow {AD} = 5\overrightarrow {DE} \)

\( \Leftrightarrow \overrightarrow {DB} + 3\overrightarrow {DC} - \overrightarrow {AD} = 5\overrightarrow {DE} \)

\( \Leftrightarrow \frac{1}{5}\overrightarrow {DB} + \frac{3}{5}\overrightarrow {DC} - \frac{1}{5}\overrightarrow {AD} = \overrightarrow {DE} \)

Vậy \(\overrightarrow {DE} = \frac{1}{5}\overrightarrow {DB} + \frac{3}{5}\overrightarrow {DC} - \frac{1}{5}\overrightarrow {AD} \).