3) 195-83; 21 14

3) $$\left(19\frac{5}{21}-8\frac{3}{14}\right)=\left(\frac{404}{21}-\frac{115}{14}\right)=\frac{404 \cdot 2}{21 \cdot 2}-\frac{115 \cdot 3}{14 \cdot 3}=\frac{808}{42}-\frac{345}{42}=\frac{808-345}{42}=\frac{463}{42}=11\frac{1}{42}$$

Ответ: $$11\frac{1}{42}$$