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

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

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

Đỗ Lê Tú Linh Sai rồi :3

Đỗ Lê Tú Linh Đúng rồi đấy 

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

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

đúng đấy bạn 

M=3/1.2-5/2.3+7/3.4-9/4.5+11/5.6-13/6.7+15/7.8+17/8.9

   =(1/1.1+2/1.2)-(2/2.3+3/2.3)+(3/3.4+4/3.4)-(4/4.5+5/4.5)+...+(8/8.9+9/8.9)(phần ... là làm tương tự nhé)

   =1/2+1-(1/3+1/2)+(1/4+1/3)-(1/5+1/4)+...+(1/9+1/8)(phần ... là làm tương tự nhé)

   =1+(1/2-1/2)+(1/3-1/3)+(1/4-1/4)+...+(1/8-1/8)-1/9

   =1+0+0+0+...+0-1/9

   =1-1/9

   =8/9

A = \(\frac{-79}{90}\)

B = \(\frac{8}{9}\)

cách giải sao chỉ mình với

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

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