Обчисли зручним способом:
Розв'язок:
1) $-2\frac{9}{14} \cdot 0{,}6 - 0{,}6 \cdot \left(-1\frac{10}{21}\right) + 0{,}6 \cdot 1\frac{1}{6} =$
$= 0{,}6 \cdot \left(-2\frac{9}{14} + 1\frac{10}{21} + 1\frac{1}{6}\right) =$
$= 0{,}6 \cdot \left(\frac{31}{21} + \frac{7}{6} - \frac{37}{14}\right) =$
$= 0{,}6 \cdot \frac{62+49-111}{42} = 0{,}6 \cdot 0 = 0$;
2) $5{,}7 \cdot 3\frac{1}{3} - 5{,}7 \cdot 4\frac{2}{3} + 4{,}7 \cdot 5\frac{1}{3} + (-4{,}7) \cdot 4 =$
$= 5{,}7 \cdot \left(3\frac{1}{3} - 4\frac{2}{3}\right) + 4{,}7 \cdot \left(5\frac{1}{3} - 4\right) =$
$= 5{,}7 \cdot \left(\frac{10}{3} - \frac{14}{3}\right) + 4{,}7 \cdot \left(\frac{16}{3} - \frac{12}{3}\right) =$
$= 5{,}7 \cdot \left(-\frac{4}{3}\right) + 4{,}7 \cdot \frac{4}{3} = \frac{4}{3} \cdot (-5{,}7 + 4{,}7) =$
$= -\frac{4}{3} = -1\frac{1}{3}$.