2.310 Вычислите: a) 4⋅(3/6-2 1/14)^2; 6) (2 1/2)^3 + 5/9 : 9/11; B) (2 1/2 - 11/14)⋅(1 3/7 + 2 -2/6)^2

a) $$4 \cdot (\frac{3}{6} - 2\frac{1}{14})^2 = 4 \cdot (\frac{1}{2} - \frac{29}{14})^2 = 4 \cdot (\frac{7}{14} - \frac{29}{14})^2 = 4 \cdot (-\frac{22}{14})^2 = 4 \cdot (-\frac{11}{7})^2 = 4 \cdot \frac{121}{49} = \frac{484}{49} = 9\frac{43}{49}$$.

б) $$(2\frac{1}{2})^3 + \frac{5}{9} : \frac{9}{11} = (\frac{5}{2})^3 + \frac{5}{9} \cdot \frac{11}{9} = \frac{125}{8} + \frac{55}{81} = \frac{125 \cdot 81 + 55 \cdot 8}{8 \cdot 81} = \frac{10125 + 440}{648} = \frac{10565}{648} = 16 \frac{197}{648}$$.

в) $$(2 \frac{1}{2} - \frac{11}{14}) \cdot (1\frac{3}{7} + 2 - \frac{2}{6})^2 = (\frac{5}{2} - \frac{11}{14}) \cdot (\frac{10}{7} + 2 - \frac{1}{3})^2 = (\frac{35}{14} - \frac{11}{14}) \cdot (\frac{30}{21} + \frac{42}{21} - \frac{7}{21})^2 = \frac{24}{14} \cdot (\frac{65}{21})^2 = \frac{12}{7} \cdot \frac{4225}{441} = \frac{4 \cdot 4225}{147} = \frac{4 \cdot 4225}{147} = \frac{16900}{147} = \frac{2414}{21} \approx 114.95$$.

Ответ: a) $$9\frac{43}{49}$$, б) $$16 \frac{197}{648}$$, в) $$114\frac{2}{21}$$.