\(a,\dfrac{\left(-11\right).3+7.\left(-11\right)}{\left(-15\right).22}=\dfrac{\left(-11\right).\left(3+7\right)}{\left(-15\right).22}=\dfrac{-11.10}{-15.22}=\dfrac{-11.2.5}{-11.3.5+2.11}=\dfrac{-1}{-3}=\dfrac{1}{3}\\ \dfrac{25.11}{15.37-15.48}\\ =\dfrac{5.5.11}{15.\left(37-48\right)}=\dfrac{5.5.11}{15.\left(-11\right)}=\dfrac{5.5.11}{3.5.\left(-11\right)}=\dfrac{5}{-3}=-\dfrac{5}{3}\\ \dfrac{-2.3.5^2}{3^2.5^3}=\dfrac{-2}{3.5}=-\dfrac{2}{15}\)
\(\dfrac{\left(-11\right).3+7.\left(-11\right)}{\left(-15\right).22}=\dfrac{-33+\left(-77\right)}{-330}=\dfrac{-110}{-330}=\dfrac{110}{330}=\dfrac{1}{3}\)
\(\dfrac{25.11}{15.37-15.48}=\dfrac{275}{15.\left(37-48\right)}=\dfrac{275}{15.\left(-11\right)}=\dfrac{275}{-165}=-\dfrac{5}{3}\)
\(\dfrac{-2.3.5^2}{3^2.5^3}=\dfrac{-2}{3.5}=-\dfrac{2}{15}\)