Respuesta

B)

Resolviendo obtenemos:

$-0,25-0,80\cdot 1,25-\left({\displaystyle \frac{-0,75}{{\displaystyle \frac{3}{4}}}}\right)=$

${\displaystyle \frac{-25}{100}}-{\displaystyle \frac{80}{100}}\cdot {\displaystyle \frac{125}{100}}-\left({\displaystyle \frac{{\displaystyle \frac{-75}{100}}}{{\displaystyle \frac{3}{4}}}}\right)=$

${\displaystyle \frac{-1}{4}}-({\displaystyle \frac{8}{10}}\cdot {\displaystyle \frac{5}{4}})-({\displaystyle \frac{-3}{4}}\cdot {\displaystyle \frac{4}{3}})=$

${\displaystyle \frac{-1}{4}}-({\displaystyle \frac{4}{5}}\cdot {\displaystyle \frac{5}{4}})-\left({\displaystyle \frac{-12}{12}}\right)=$

${\displaystyle \frac{-1}{4}}-\left({\displaystyle \frac{20}{20}}\right)+1=$

${\displaystyle \frac{-1}{4}}-1+1={\displaystyle \frac{-1}{4}}$