\(-{5\over6}\)

\(-256\over35\)

We have, \({-2\over3}\times{3\over5}+{5\over2}-{3\over5}\times{1\over6}={-2\over3}\times{3\over5}-{3\over5}\times{1\over6}+{5\over2}\) [by associativity]
\(={3\over5}\times{-2\over3}+{3\over5}\times{-1\over6}+{5\over2}\)         [by commutativity]
\(={3\over5}\times({-2\over3}-{1\over6})+{5\over2}\)
[by distributivity, taking \({3\over5}\)  as common factor]
\(={3\over5}\times({-4-1\over6})+{5\over2} \ \ \ \ [\because LCM \ of \ 3 \ and \ 6=6] \)
\(\\ ={3\over5}\times({-5\over6})+{5\over2}={3\over5}\times {-5\over6}+{5\over2}={-1\over2}+{5\over2}={-1+5\over2}\)
\(\\ ={4\over2}=2\)

\(\frac{2}{3} \times (\frac{-6}{7} \times \frac{4}{5})=\frac{2}{3} \times \frac{-24}{35}\times = \frac{-48}{105} \)
\((\frac{2}{3} \times \frac{-6}{7}) \times \frac{4}{5}=\frac{-12}{21} \times \frac{4}{5}\times = \frac{-48}{105} \)
So, Yes; \(\frac{2}{3} \times (\frac{-6}{7} \times \frac{4}{5})=(\frac{2}{3} \times \frac{-6}{7})\times \frac{4}{5}\).

Here,\({1\over3}\times({6\times {4\over3}}) as ({1\over3}\times 6)\times{4\over3} \) and \(({1\over3}\times6)\times{4\over3}={6\over3}\times{4\over3}={8\over3}\)
\(\therefore {1\over3} \times({6 \times {4\over3}})=({1\over3}\times 6)\times {4\over3}\)
Hence, we use the associative property for rational numbers.

we have, \(19\over-6\) so, additive inverse of \(19\over-6\) is \(19\over6\)

\(-{2\over6},-{1\over6},0,{1\over6},{2\over6}\)