Ответ:

  1. а) \(\frac{2}{3}\sqrt{12,96}+\frac{1}{7}\sqrt{4,41}=\frac{2}{3}\cdot3,6+\frac{1}{7}\cdot2,1=2,4+0,3=2,7\).
  2. б) \(\left(-\frac{2}{\sqrt{6}}\right)^2=\frac{4}{6}=\frac{2}{3}\).
  3. в) \(1\frac{40}{81}=\frac{121}{81}\), поэтому \(\sqrt{1\frac{40}{81}\cdot\frac{4}{49}}-\sqrt{1}=\sqrt{\frac{484}{3969}}-1=\frac{22}{63}-1=-\frac{41}{63}\).