Răspuns:
Regula semnelor
[tex]\boxed{\boxed{\boldsymbol{(+)\cdot(-)=(-)}} \ \boxed{\boldsymbol{(-)\cdot(+)=(-)}} \ \boxed{\boldsymbol{(-)\cdot(-)=(+)}} \ \boxed{\boldsymbol{(+)\cdot(+)=(+)}}}[/tex]
[tex]b) = -\dfrac{3}{41} \cdot \dfrac{40 + 1}{2} - \dfrac{8}{3} \cdot \bigg(1 - \dfrac{5}{\not13} \cdot \dfrac{\not13}{2}\bigg) = \\[/tex]
[tex]= -\dfrac{3}{\not41} \cdot \dfrac{\not41}{2} - \dfrac{8}{3} \cdot \bigg(^{2)} 1 - \dfrac{5}{2}\bigg) = -\dfrac{3}{2} - \dfrac{8}{3} \cdot \bigg(\dfrac{2}{2} - \dfrac{5}{2}\bigg) = -\dfrac{3}{2} - \dfrac{8}{3} \cdot \bigg(- \dfrac{5 - 2}{2}\bigg)\\[/tex]
[tex]= -\dfrac{3}{2} - \dfrac{8}{3} \cdot \bigg(- \dfrac{5 - 2}{2}\bigg) = -\dfrac{3}{2} - \dfrac{8}{3} \cdot \bigg(- \dfrac{3}{2}\bigg) = -\dfrac{3}{2} + \dfrac{8}{\not3} \cdot \dfrac{\not3}{2} = -\dfrac{3}{2} + \dfrac{8}{2}\\[/tex]
[tex]= \dfrac{8 - 3}{2} = \bf\dfrac{5}{2} = 2 \dfrac{1}{2}[/tex]
[tex]c) = \bigg(-\dfrac{9}{13}\bigg) \cdot \bigg( - \dfrac{39}{99}^{3)} \bigg) \cdot \bigg(-\dfrac{20}{21}\bigg) = \bigg(-\dfrac{9}{13}\bigg) \cdot \bigg( - \dfrac{13}{33}\bigg) \cdot \bigg(-\dfrac{20}{21}\bigg) = \\[/tex]
[tex]= -\dfrac{\not9}{\not13} \cdot \dfrac{\not13}{\not33} \cdot \dfrac{20}{21} = -\dfrac{\not3}{1} \cdot \dfrac{1}{11} \cdot \dfrac{20}{\not21} = -\dfrac{1}{1} \cdot \dfrac{1}{11} \cdot \dfrac{20}{7} \\[/tex]
[tex]= \bf-\dfrac{20}{77}[/tex]
[tex]d) \ \bigg(-\dfrac{4}{5}\bigg) \cdot \bigg(- \dfrac{3}{16} + \dfrac{^{2)} 1}{8} \bigg) + \dfrac{3}{10} \cdot \bigg(^{2)} 1 - \dfrac{1}{2}\bigg) =\\[/tex]
[tex]= \bigg(-\dfrac{4}{5}\bigg) \cdot \bigg(- \dfrac{3}{16} + \dfrac{2}{16} \bigg) + \dfrac{3}{10} \cdot \bigg(\dfrac{2}{2} - \dfrac{1}{2}\bigg) = \bigg(-\dfrac{4}{5}\bigg) \cdot \bigg(- \dfrac{3 - 2}{16}\bigg) + \dfrac{3}{10} \cdot \bigg(\dfrac{2 - 1}{2}\bigg)\\[/tex]
[tex]= \bigg(-\dfrac{4}{5}\bigg) \cdot \bigg(- \dfrac{1}{16}\bigg) + \dfrac{3}{10} \cdot \bigg(\dfrac{1}{2}\bigg) = \dfrac{\not4}{5} \cdot \dfrac{1}{\not16} + \dfrac{3}{10} \cdot \dfrac{1}{2} = \dfrac{1}{20} + \dfrac{3}{20} = \dfrac{4}{20}^{(4} \\[/tex]
[tex]=\bf \dfrac{1}{5}[/tex]
