Выполним задание:
$$√{a^2+8ab+16b^2} = √{(3\frac{3}{7})^2 + 8 \cdot 3\frac{3}{7} \cdot \frac{1}{7} + 16 \cdot (\frac{1}{7})^2}$$
$$3\frac{3}{7} = \frac{3 \cdot 7 + 3}{7} = \frac{21 + 3}{7} = \frac{24}{7}$$
$$√{(\frac{24}{7})^2 + 8 \cdot \frac{24}{7} \cdot \frac{1}{7} + 16 \cdot (\frac{1}{7})^2} = √{\frac{576}{49} + \frac{192}{49} + \frac{16}{49}} = √{\frac{576 + 192 + 16}{49}} = √{\frac{784}{49}} = \frac{√{784}}{√{49}} = \frac{28}{7} = 4$$
Ответ: 4