a)

\(\Rightarrow A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)

\(\Rightarrow A=\frac{1}{5}+\frac{2}{7}\)

\(\Rightarrow A=\frac{17}{35}\)

b)

\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{56}-\frac{1}{61}\right)\)

\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{61}\right)\)

\(\Rightarrow B=5.\frac{50}{671}=\frac{250}{671}\)

c)

\(\Rightarrow C=1-\left(\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+....+\frac{1}{49.25}\right)\)

\(\Rightarrow C=1-2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{49.50}\right)\)

\(\Rightarrow C=1-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\right)\)

\(\Rightarrow C=1-1-\frac{1}{25}\)

\(\Rightarrow C=\frac{1}{25}\)

1+1=3 :)))

a)

\(\begin{array}{l}1\frac{1}{2} + \frac{1}{5}.\left[ {\left( { - 2\frac{5}{6} + \frac{1}{3}} \right)} \right]\\ = \frac{3}{2} + \frac{1}{5}.\left[ {\left( { - \frac{{17}}{6} + \frac{2}{6}} \right)} \right]\\ = \frac{3}{2} + \frac{1}{5}.\frac{{ - 15}}{6}\\ = \frac{3}{2} + \frac{{ - 1}}{2}\\ = \frac{2}{2}\\=1\end{array}\)      

b)

\(\begin{array}{l}\frac{1}{3}.\left( {\frac{2}{5} - \frac{1}{2}} \right):{\left( {\frac{1}{6} - \frac{1}{5}} \right)^2}\\ = \frac{1}{3}.\left( {\frac{4}{{10}} - \frac{5}{{10}}} \right):{\left( {\frac{5}{{30}} - \frac{6}{{30}}} \right)^2}\\ = \frac{1}{3}.\frac{{ - 1}}{{10}}:{\left( {\frac{{ - 1}}{{30}}} \right)^2}\\ = \frac{{ - 1}}{{30}}:\frac{1}{{{{30}^2}}}\\ = \frac{{ - 1}}{{30}}{.30^2}\\ =  - 30\end{array}\)

\(1)A=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\)

\(=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}\)

\(=\frac{2}{4}=\frac{1}{2}\)

\(2)B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)

\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.\frac{4.4}{4.5}\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)

\(=\frac{1.2.3.4}{2.3.4.5}=\frac{1}{5}\)

\(3)C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)

\(=\frac{2.2.3.3.4.4.5.5}{1.3.2.4.3.5.4.6}\)

\(=\frac{2.5}{1.6}=\frac{2.5}{1.3.2}=\frac{5}{3}\)

\(4)D=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{30}\right)\)

\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{6}{30}-\frac{5}{30}-\frac{1}{30}\right)\)

\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right).0=0\)

\(5)M=8\frac{2}{7}-\left(3\frac{4}{9}+3\frac{9}{7}\right)\)               \(N=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\)

\(=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)\)                         \(=\left(\frac{92}{9}+\frac{13}{5}\right)-\frac{56}{9}\)

\(=\frac{58}{7}-\left(\frac{217}{63}+\frac{270}{63}\right)\)                     \(=\left(\frac{460}{45}+\frac{117}{45}\right)-\frac{280}{45}\)

\(=\frac{58}{7}-\frac{487}{63}\)                                          \(=\frac{577}{45}-\frac{280}{45}\)

\(=\frac{522}{63}-\frac{487}{63}=\frac{5}{9}\)                             \(=\frac{33}{5}\)

\(P=M-N\)

\(\Rightarrow P=\frac{5}{9}-\frac{33}{5}\)

\(\Rightarrow P=\frac{25}{45}-\frac{297}{45}\)

\(\Rightarrow P=\frac{-272}{45}\)

Vậy P = \(\frac{-272}{45}\)

\(6)E=10101\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)

\(=\frac{5}{11}+\frac{5}{22}-\left(10101.\frac{4}{111111}\right)\)

\(=\frac{10}{22}+\frac{5}{22}-\frac{4}{11}\)

\(=\frac{15}{22}-\frac{8}{22}=\frac{7}{22}\)

\(7)F=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{64}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)

\(=\frac{1\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}{2\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}.\frac{3\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{256}+\frac{1}{64}\right)}{1\left(1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}\right)}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{3\left(\frac{16}{64}-\frac{4}{64}+\frac{1}{64}-\frac{1}{256}\right)}{1\left(\frac{64}{64}-\frac{16}{64}+\frac{4}{64}-\frac{1}{64}\right)}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{3\left(\frac{13}{64}-\frac{1}{256}\right)}{1.\frac{51}{64}}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{3\left(\frac{52}{256}-\frac{1}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{3\left(\frac{51}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{\frac{153}{256}}{\frac{51}{64}}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{153}{256}:\frac{51}{64}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{3}{4}+\frac{5}{8}\)

\(=\frac{3}{8}+\frac{5}{8}=1\)

Xin lỗi tớ đã làm hết buổi tối mà chỉ có 7 bài mong bạn thông cảm cho mình nhé !

1.\(A=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\)

\(\Rightarrow A=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}\)

\(\Rightarrow A=\frac{2}{4}=\frac{1}{2}\)

\(\frac{5}{7}\times\frac{1}{3}-\frac{5}{7}\times\frac{1}{4}-\frac{5}{7}\times\frac{1}{2}\)

\(=\frac{5}{7}\times\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{2}\right)\)

\(=\frac{5}{7}\times\left(\frac{4}{12}-\frac{3}{12}-\frac{6}{12}\right)\)

\(=\frac{5}{7}\times\left(\frac{4-3-6}{12}\right)\)

\(=\frac{5}{7}\times\frac{-5}{12}\)

\(=\frac{5\times\left(-5\right)}{7\times12}\)

\(=\frac{-25}{84}\)

\(\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{2}\)

= \(\frac{5}{7}.\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{2}\right).1\)

\(=\frac{5}{7}.\frac{-5}{12}\)

\(=-\frac{25}{84}\)

\(75\%-1\frac{1}{2}+0,5:\frac{5}{12}-\left(\frac{-1}{2}\right)^2\)

= \(\frac{3}{4}-\frac{3}{2}+\frac{1}{2}:\frac{5}{12}-\frac{1}{4}\)

\(\frac{3}{4}-\frac{3}{2}+\frac{1}{2}.\frac{5}{12}-\frac{1}{4}\)

\(\frac{3}{4}-\frac{3}{2}+\frac{5}{24}-\frac{1}{4}\)

\(\frac{18}{24}-\frac{36}{24}+\frac{5}{24}-\frac{6}{24}\)

-19/24