\(\left(-1-\frac{1}{12}\right).\left(-1-\frac{1}{13}\right).\left(-1-\frac{1}{14}\right)...\left(-1-\frac{1}{2017}\right)\)

\(=\frac{-13}{12}.\frac{-14}{13}.\frac{-15}{14}...\frac{-2018}{2017}\)

\(=\frac{-13}{12}.\frac{14}{-13}.\frac{-15}{14}...\frac{2018}{-2017}\)

\(=\frac{\left(-13\right).14.\left(-15\right)...2018}{12.\left(-13\right).14...2017}=\frac{2018}{12}=\frac{1009}{6}\)

\(\frac{-3}{5}.\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)\)

=\(\frac{-3}{5}.\)\(\frac{14}{7}\)

=\(-\frac{3}{5}.2\)

=-6/5

\(15\times\frac{1}{4}\times\frac{-5}{7}-25\times\frac{1}{4}\times\frac{-5}{7}\)

\(=\frac{1}{4}\times\frac{-5}{7}\left(15-25\right)\)

\(=\frac{1}{4}\times\frac{-5}{7}\times\left(-10\right)\)

\(=\frac{25}{14}\)

\(15\times\frac{1}{4}\times\frac{-5}{7}-25\times\frac{1}{4}\times\frac{-5}{7}=\frac{1}{4}\times\frac{-5}{7}\left(15-25\right)=\frac{1}{4}\times\frac{-5}{7}\times\left(-10\right)=\frac{25}{14}\)

M=3.(\(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-....+\frac{1}{59}-\frac{1}{60}\)\(\frac{1}{61}\))

M= 3.(\(\frac{1}{5}-\frac{1}{61}\))

M=\(\frac{168}{305}\)

\(M=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)

\(M=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)

\(M=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)\)

\(M=\frac{84}{305}\)

\(=-\frac{3}{4}\left(5\frac{3}{13}+\frac{36}{13}\right)\)

\(=-\frac{3}{4}.\left(5+3\right)\)

\(=-\frac{3}{4}.8\)

\(=-6\)

đặt A = \(\frac{1}{2^2}+\frac{2}{2^3}+...+\frac{10}{2^{11}}\)

2A = \(\frac{1}{2}+\frac{2}{2^2}+...+\frac{10}{2^{10}}\)

2A - A = ( \(\frac{1}{2}+\frac{2}{2^2}+...+\frac{10}{2^{10}}\)) - ( \(\frac{1}{2^2}+\frac{2}{2^3}+...+\frac{10}{2^{11}}\))

A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}-\frac{10}{2^{11}}\)

A = \(1-\frac{1}{2^{10}}-\frac{10}{2^{11}}\)

       \(\left(\frac{42}{32}:\frac{7}{8}\right).\frac{3}{4}\)

\(=\left(\frac{42}{32}.\frac{8}{7}\right).\frac{3}{4}\)

\(=\frac{3}{2}.\frac{3}{4}\)

\(=\frac{9}{8}\)