Đặt biểu thức trên là A

\(\Rightarrow A=7x\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\right)\)

Đặt biểu thức trong ngoặc là B

\(\Rightarrow B=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{9x10}+\frac{1}{10x11}\)

Đây là dạng tính tổng các phân số mà mỗi phân số có:

-Tử số là hiệu của hai thừa số ở mẫu

-Mẫu số của phân số liền sau là tích của hai thừa số mà thừa số thứ nhất là thừa số thứ hai ở mẫu của phân số liền trước

\(B=\frac{2-1}{1x2}+\frac{3-2}{2x3}+\frac{4-3}{3x4}+...+\frac{10-9}{9x10}+\frac{11-10}{10x11}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

\(B=1-\frac{1}{11}=\frac{10}{11}\Rightarrow A=\frac{70}{11}\)

\(\frac{7}{2}+\frac{7}{6}+\frac{7}{12}+\frac{7}{20}+\frac{7}{30}+\frac{7}{42}+\frac{7}{56}+\frac{7}{72}+\frac{7}{90}\)\(\frac{7}{90}\)

=\(\frac{7}{2+6+12+20+30+42+56+72+90}\)

=\(\frac{63}{10}\)

=6.3

hơi lệch

\(1,A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\\ 2A=\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{99}-\dfrac{1}{103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{103}=\dfrac{100}{309}\\ A=\dfrac{100}{309}\cdot\dfrac{1}{2}=\dfrac{50}{309}\)

\(2,A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\\ A=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\right)\\ A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ A=7\left(1-\dfrac{1}{10}\right)=7\cdot\dfrac{9}{10}=\dfrac{63}{10}\)

Bài 1: 

Ta có: \(A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\)

\(=\dfrac{1}{2}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{100}{309}=\dfrac{50}{309}\)

Bài 2: 

Ta có: \(A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\)

\(=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\)

\(=7\left(1-\dfrac{1}{10}\right)\)

\(=\dfrac{63}{10}\)

A = 1 - \(\dfrac{5}{6}\)+\(\dfrac{7}{12}\)-\(\dfrac{9}{20}\)+\(\dfrac{11}{30}\)-\(\dfrac{13}{42}\)+\(\dfrac{15}{56}\) - \(\dfrac{17}{72}\)+\(\dfrac{19}{90}\)+\(\dfrac{23}{132}\)-\(\dfrac{25}{156}\)

A = 1 - \(\dfrac{5}{2.3}\)+\(\dfrac{7}{3.4}\)-\(\dfrac{9}{4.5}\)+\(\dfrac{11}{5.6}\)-\(\dfrac{13}{6.7}\)+\(\dfrac{15}{7.8}\)-\(\dfrac{17}{8.9}\)+\(\dfrac{19}{9.10}\)+\(\dfrac{23}{11.12}\)-\(\dfrac{25}{12.13}\)

A = 1 - \(\dfrac{1}{2}-\dfrac{1}{3}\)+\(\dfrac{1}{3}+\dfrac{1}{4}\)-\(\dfrac{1}{4}-\dfrac{1}{5}\)+...+\(\dfrac{1}{11}+\dfrac{1}{12}\)- \(\dfrac{1}{12}-\dfrac{1}{13}\)

A = 1 - \(\dfrac{1}{2}\) - \(\dfrac{1}{13}\) 

A = \(\dfrac{1}{2}\) - \(\dfrac{1}{13}\)

A = \(\dfrac{11}{26}\) 

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