\(\frac{7}{9}\)

          \(\frac{1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}}{4+\frac{4}{7}+\frac{4}{7^2}-\frac{4}{7^3}}.\frac{858585}{313131}.\left(-1\frac{14}{17}\right)\)\

\(=\frac{1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}}{4.\left(1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}\right)}.\frac{85}{31}.\left(-\frac{31}{17}\right)\)

\(=\frac{1}{4}.\frac{85}{31}.\left(-\frac{31}{17}\right)\)

\(=-\frac{5}{4}\)

\(=\frac{1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}}{4\left(1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}\right)}.\frac{85.10101}{31.10101}.\left(-\frac{31}{17}\right).\)

\(=\frac{1}{4}.\frac{3.17}{31}.\left(-\frac{31}{17}\right)=-\frac{3}{4}\)

\(P=\frac{\frac{3}{7}-\frac{3}{13}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)

\(=\frac{3\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)

\(=\frac{3}{5}+\frac{1}{-7}\)

\(=\frac{16}{35}\)

\(A=\frac{3.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{1.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}{\left(-7\right).\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)

\(A=\frac{3}{5}+\frac{-1}{7}\)

\(A=\frac{21}{35}+\frac{-5}{35}\)

\(A=\frac{16}{35}\)

\(A=\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)

\(=\frac{3.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-\frac{7}{2}+\frac{7}{3}-\frac{7}{4}+\frac{7}{5}}\)

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

=3/5+(1/-7)

=3/5-1/7

=16/35

Đấm zô đúng thì biết câu trả lời của mình hay lắm đó cứ thử đi

\(=\frac{12.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{3.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{7.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\)

\(=\frac{12}{4}:\frac{3}{7}=3.\frac{7}{3}=7\)

=7/19 nhé do tien toan

\(A=\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)

\(=\frac{3\left(\frac{1}{7}-\frac{1}{17}-\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}-\frac{1}{37}\right)}+\frac{1.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)

\(=\frac{3}{5}+\frac{-1}{7}\)

\(=\frac{21}{35}-\frac{5}{35}\)

\(=\frac{16}{35}\)

\(A=\frac{3.\left(\frac{1}{7}-\frac{1}{17}-\frac{1}{37}\right)}{5.\left(\frac{1}{7}-\frac{1}{17}-\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{7.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)

\(A=\frac{3}{5}+\frac{1}{7}=\frac{21}{35}+\frac{5}{35}=\frac{26}{35}\)

5và 3/8-1 và 5/6

cách này mình tự nghĩ 

\(\hept{\begin{cases}A=\frac{4}{7}+5+\frac{3}{7^2}+\frac{5}{7^3}+\frac{6}{7^4}\\B=\frac{5}{7^4}+5+\frac{6}{7^2}+\frac{4}{7}+\frac{5}{7^3}\end{cases}}\)

\(\Rightarrow A-B=\left(\frac{4}{7}-\frac{4}{7}\right)+\left(\frac{5}{7^3}-\frac{5}{7^3}\right)+\left(5-5\right)+\left(\frac{3}{7^2}-\frac{6}{7^2}\right)+\left(\frac{6}{7^4}-\frac{5}{7^4}\right)\)

\(\Rightarrow A-B=-\frac{3}{7^2}+\frac{1}{7^4}\)

\(\Rightarrow A-B=\frac{-3\times7^2}{7^4}+\frac{1}{7^4}\)

mà \(-3\times7^2< 1\Rightarrow\frac{1}{7^4}>\frac{-3\times7^2}{7^4}\Rightarrow B>A\)

\(a,15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\\ =15\frac{3}{13}-3\frac{4}{7}-8\frac{3}{13}\\ =7\frac{3}{13}-3\frac{4}{7}\\ =\frac{94}{13}-\frac{25}{7}\\ =\frac{94\cdot7-25\cdot13}{13\cdot7}\\ =\frac{333}{91}\)

\(b,\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\\ =7\frac{4}{9}-3\frac{4}{9}+4\frac{7}{11}\\ =4\frac{4}{9}+4\frac{7}{11}\\ =\frac{40}{9}+\frac{51}{11}\\ =\frac{40\cdot11+51\cdot9}{11\cdot9}\\ =\frac{899}{99}\)

\(c,=-\frac{7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}\right)+\frac{52}{9}\\ =-\frac{7}{9}\cdot1+\frac{52}{9}\\ =\frac{-7+52}{9}\\ =\frac{45}{9}=5\)

\(d,=\frac{50}{100}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{75}{100}\\ =\frac{50\cdot2\cdot2\cdot2\cdot5\cdot7\cdot25\cdot3}{50\cdot2\cdot3\cdot5\cdot7\cdot25\cdot2\cdot2}=1\)

\(e,=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\\ =1-\frac{1}{43}\\ =\frac{42}{43}\)

a) Ta có: \(15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\)

\(=15+\frac{3}{13}-3-\frac{4}{7}-8-\frac{3}{13}\)

\(=4-\frac{4}{7}=\frac{24}{7}\)

b) Ta có: \(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)

\(=7+\frac{4}{9}+4+\frac{7}{11}-3-\frac{4}{9}\)

\(=8+\frac{7}{11}=\frac{95}{11}\)

c) Ta có: \(\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+5\frac{7}{9}\)

\(=\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+\frac{-7}{9}\cdot\frac{-52}{7}\)

\(=\frac{-7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}-\frac{52}{7}\right)\)

\(=\frac{-7}{9}\cdot\frac{45}{-7}=5\)

d) Ta có: \(50\%\cdot1\frac{1}{3}\cdot10\cdot\frac{7}{35}\cdot0.75\)

\(=\frac{1}{2}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{3}{4}\)

\(=5\cdot\frac{7}{35}=1\)

e) Ta có: \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

\(=1-\frac{1}{43}=\frac{43}{43}-\frac{1}{43}\)

\(=\frac{42}{43}\)