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

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

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

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

\(\frac37\) x (\(x\) - \(\frac{14}{9}\)) = - \(\frac{11}{7}\) x (\(x+\frac{14}{11}\))

\(\frac37x\) - \(\frac23\) = - \(\frac{11}{7}x\) - 2

\(\frac37\)\(x\) + \(\frac{11}{7}\)\(x\) = - 2 + \(\frac23\)

2\(x\) = \(-\frac43\)

\(x\) = - \(\frac43:2\)

\(x\) = - \(\frac23\)

Vậy \(x=-\frac23\)

73​ x (\(x\) - \(\frac{14}{9}\)) = - \(\frac{11}{7}\) x (\(x + \frac{14}{11}\))

\(\frac{3}{7} x\) - \(\frac{2}{3}\) = - \(\frac{11}{7} x\) - 2

\(\frac{3}{7}\)\(x\) + \(\frac{11}{7}\)\(x\) = - 2 + \(\frac{2}{3}\)

2\(x\) = \(- \frac{4}{3}\)

\(x\) = - \(\frac{4}{3} : 2\)

\(x\) = - \(\frac{2}{3}\)

Vậy \(x = - \frac{2}{3}\)

\(A=\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)

\(A=\frac{-4}{5}+\frac{4}{3}+\frac{-5}{4}+\frac{14}{5}-\frac{7}{3}\)

\(A=\left(\frac{-4}{5}+\frac{14}{5}\right)+\left(\frac{4}{3}-\frac{7}{3}\right)+\frac{-5}{4}\)

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

\(A=\frac{-1}{4}\)

A=\(\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)\(\frac{7}{3}\)

  =\(\frac{-4}{5}+\frac{4}{3}+\frac{-5}{4}+\frac{14}{5}+\frac{-7}{3}\)=\(\left(\frac{-4}{5}+\frac{14}{5}\right)+\left(\frac{4}{3}+\frac{-7}{3}\right)+\frac{-5}{4}\)

  =\(\frac{10}{5}+\frac{-3}{3}+\frac{-5}{4}\)=\(2-1+\frac{-5}{4}\)=\(1+\frac{-5}{4}\)=\(\frac{4}{4}+\frac{-5}{4}\)=\(\frac{4-5}{4}\)=\(\frac{-1}{4}\)

Ta có \(\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}=\left(\frac{3}{10}-\frac{6}{20}-\frac{4}{15}-\frac{1}{20}\right).\frac{5}{19}=-\left(\frac{4}{15}+\frac{1}{20}\right).\frac{5}{19}\)

\(=-\frac{1}{5}\left(\frac{4}{3}+\frac{1}{4}\right).\frac{5}{19}=-\left(\frac{4}{3}+\frac{1}{4}\right).\frac{1}{19}=-\frac{19}{12}.\frac{1}{19}=-\frac{1}{12}\)

Lại có \(\left(\frac{1}{14}+\frac{1}{7}-\frac{3}{35}\right).\frac{-4}{3}=\left(\frac{1}{14}+\frac{1}{7}-\frac{5}{35}+\frac{2}{35}\right).\frac{-4}{3}=\left(\frac{1}{14}+\frac{2}{35}\right).\frac{-4}{3}\)

\(=\frac{1}{7}\left(\frac{1}{2}+\frac{2}{5}\right).\frac{-4}{3}=\left(\frac{1}{2}+\frac{2}{5}\right).\frac{-4}{21}=\frac{7}{10}.\frac{-4}{21}=\frac{-2}{15}\)

Khi đó \(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{3}{35}\right).\frac{-4}{3}}=\frac{\frac{1}{12}}{-\frac{2}{15}}=-\frac{5}{8}\)

Làm lại : 

\(\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\left(-\frac{4}{3}\right)=\left(\frac{1}{14}+\frac{1}{7}-\frac{5}{35}+\frac{2}{35}\right)\left(-\frac{4}{3}\right)=\left(\frac{1}{14}+\frac{2}{35}\right)\left(-\frac{4}{3}\right)\)

\(=\frac{1}{7}\left(\frac{1}{2}+\frac{2}{5}\right).\frac{-4}{3}=\frac{-4}{21}.\frac{9}{10}=\frac{-6}{35}\)

Khi đó A = \(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}=\frac{-\frac{1}{12}}{-\frac{6}{35}}=\frac{35}{72}\)

ick mình

Phải làm đúng mới k

trả lòi bằng 1  tick minh nha

2) 1\26+1\27+1\28+........+1\50=1+1\2+1\3+......+1\50 -( 1+1\2+1\3+.....+1\25)=1+1\2+1\3+....+1\50-2.(1\2+1\4+1\6+....+1\50)=1-1\2+1\3-1\4+.....+1\49-1\50=vế phải(đpcm)