\(S=\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{42}\)

\(=1+\frac{1}{2}-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)+ \(\frac{1}{6}-\frac{1}{6}-\frac{1}{7}+\frac{1}{7}\)+ \(\frac{1}{8}-\frac{1}{8}-\frac{1}{9}\)

\(=1-\frac{1}{9}\)

\(=\frac{9}{9}-\frac{1}{9}\)

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

Chúc bạn học tốt !!! 

Telex à bn?

Bn xem Telex của bn có bị lỗi ko nhé.

ko đâu bn

A = \(\frac{-79}{90}\)

B = \(\frac{8}{9}\)

cách giải sao chỉ mình với

\(=\left(1+\frac{1}{2}\right)-1+\frac{1}{6}+\left(\frac{1}{2}+\frac{1}{12}\right)-\frac{1}{2}+\frac{1}{20}+\left(\frac{1}{3}+\frac{1}{30}\right)-\frac{1}{3}+\frac{1}{42}+\left(\frac{1}{4}+\frac{1}{56}\right)-\frac{1}{4}+\frac{1}{72}\)

=\(=\left(1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{72}\right)\)

\(=0+\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\right)=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}=\left(1-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...+\left(\frac{1}{8}-\frac{1}{8}\right)\)\(=\left(\frac{9}{9}-\frac{1}{9}\right)+0+...+0=\frac{8}{9}\)

Đỗ Lê Tú Linh Sai rồi :3

Đỗ Lê Tú Linh Đúng rồi đấy 

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

\(\frac{8}{9}\) k nha

ai k mk mk k lại

Đáp án là 8/9 nhé

dễ mà bn

\(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}\)

\(=\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}\)

\(=\left(1+\frac{1}{2}\right)-\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)-\left(\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{5}+\frac{1}{6}\right)-\left(\frac{1}{6}+\frac{1}{7}\right)\)

\(+\left(\frac{1}{7}+\frac{1}{8}\right)-\left(\frac{1}{8}+\frac{1}{9}\right)\)

\(=1+\frac{1}{2}-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-\frac{1}{4}-\frac{1}{5}+\frac{1}{5}+\frac{1}{6}-\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+\frac{1}{8}-\frac{1}{8}-\frac{1}{9}\)

\(=1-\frac{1}{9}=\frac{8}{9}\)

thanks

\(A=\frac{3}{2}-\frac{5}{6}+\frac{13}{12}-\frac{19}{20}+\frac{31}{30}-\frac{41}{42}+\frac{57}{56}-\frac{71}{72}+\frac{91}{90}-\frac{109}{110}\)

\(\Rightarrow A=\left(1+\frac{1}{2}\right)-\left(1-\frac{1}{6}\right)+\cdot\cdot\cdot+\left(1+\frac{1}{90}\right)-\left(1-\frac{1}{110}\right)\)

\(\Rightarrow A=1+\frac{1}{2}-1+\frac{1}{6}+\cdot\cdot\cdot+1+\frac{1}{90}-1+\frac{1}{110}\)

\(\Rightarrow A=\left[\left(1-1\right)+\frac{1}{2}+\frac{1}{6}\right]+\cdot\cdot\cdot+\left[\left(1-1\right)+\frac{1}{90}+\frac{1}{110}\right]\)

\(\Rightarrow A=\frac{1}{2}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{90}+\frac{1}{110}\)

\(\Rightarrow A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{10}-\frac{1}{11}\)

\(\Rightarrow A=1-\frac{1}{11}\)

\(\Rightarrow A=\frac{10}{11}\)