a) (2\frac{1}{3}a - \frac{1}{14}b )^2; B) (-1,2x-4\frac{1}{6}y )^2;

a)

${(2\frac{1}{3}a-\frac{1}{14}b)}^2={\left(\frac{7}{3}a\right)}^2-2·\frac{7}{3}a·\frac{1}{14}b+{\left(\frac{1}{14}b\right)}^2=\frac{49}{9}{a}^2-\frac{1}{3}ab+\frac{1}{196}{b}^2$

B)

${(-\mathrm{1,2}x-4\frac{1}{6}y)}^2={(-\mathrm{1,2}x)}^2+2·(-\mathrm{1,2}x)·(-4\frac{1}{6}y)+{(-4\frac{1}{6}y)}^2=\mathrm{1,44}{x}^2+10xy+\frac{625}{36}{y}^2$

Ответ: a) $$\frac{49}{9}a^2 - \frac{1}{3}ab + \frac{1}{196}b^2$$; B) $$1,44x^2 + 10xy + \frac{625}{36}y^2$$