\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{110}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)

mk giống bạn lê minh anh

Đặt A = \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{90}+\frac{1}{110}\)

A=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{9.10}+\frac{1}{10.11}\)

A=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

A=\(\frac{1}{2}-\frac{1}{11}\)

A=\(\frac{9}{22}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.......+\frac{1}{90}+\frac{1}{110}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...........+\frac{1}{9.10}+\frac{1}{10.11}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...........+\frac{1}{10}-\frac{1}{11}\)

\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\)

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

\(=\frac{1}{2}-\frac{1}{11}\)

\(=\frac{9}{22}\)

Tổng quát: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+..+\frac{1}{110}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{10.11}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{10}-\frac{1}{11}\)

\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)

Ta có:  \(\frac{1}{6}+\frac{1}{12}+....+\frac{1}{110}\)

    \(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{10.11}\)

    \(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\)

    \(\Rightarrow\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+........+\frac{1}{110}\)

\(=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+.......+\frac{1}{10\times11}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{10}-\frac{1}{11}\)

\(=\frac{1}{2}-\frac{1}{11}\)

\(=\frac{11}{22}-\frac{2}{22}\)

\(=\frac{9}{22}\)

Ta viết lại biểu thức A như sau:

\(A=-\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{11.12}\right)\)

\(A=-\left(\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+...+\dfrac{12-11}{11.12}\right)\)

\(A=-\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{12}\right)\)

\(A=-\left(\dfrac{1}{4}-\dfrac{1}{12}\right)\)

\(A=-\dfrac{1}{6}\)

\(B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{90}+\frac{1}{110}\)

\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{9.10}+\frac{1}{10.11}\)

\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{10}\)\(+\frac{1}{10}-\frac{1}{11}\)

\(B=\frac{1}{2}-\frac{1}{11}\)

\(B=\frac{9}{22}\)

Thôi chết! Ở hàng thứ 3 của bài Mình có ghi 1/2 - 1/3 + 1/3 - 1/4 +1/4 - 1/5 .... nha bạn! Không phải bằng đâu ^.^

\(\dfrac{1}{20}=\dfrac{1}{4x5}=\dfrac{1}{4}-\dfrac{1}{5}\)

Tương tự các phân số khác

S= \(\dfrac{1}{4}-\dfrac{1}{12}=\dfrac{1}{6}\)

\(\dfrac{1}{20}+\dfrac{1}{30}\)+ \(\dfrac{1}{42}\)+\(\dfrac{1}{56}\)+\(\dfrac{1}{72}\)+\(\dfrac{1}{90}\)+\(\dfrac{1}{110}\)+\(\dfrac{1}{132}\)

= \(\dfrac{1}{4\times5}\)+\(\dfrac{1}{5\times6}\)+\(\dfrac{1}{6\times7}\)+\(\dfrac{1}{7\times8}\)+\(\dfrac{1}{8\times9}\)+\(\dfrac{1}{9\times10}\)+\(\dfrac{1}{10\times11}\)+\(\dfrac{1}{11\times12}\)

= \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+\(\dfrac{1}{9}\)-\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)-\(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{12}\)

= \(\dfrac{1}{4}\) - \(\dfrac{1}{12}\)

= \(\dfrac{3}{12}\) - \(\dfrac{1}{12}\)

= \(\dfrac{2}{12}\)

=\(\dfrac{1}{6}\)

=1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10+1/10x11+1/11x12

=1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11+1/11-1/12

=1/4-1/12=3/12-1/12=2/12=1/6

`=1/[4xx5]+1/[5xx6]+1/[6xx7]+...+1/[11xx12]`

`=1/4-1/5+1/5-1/6+1/6-1/7+...+1/11-1/12`

`=1/4-1/12=3/12-1/12=2/12=1/6`

\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\\ =\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}+\dfrac{1}{8\times9}+\dfrac{1}{9\times10}+\dfrac{1}{10\times11}+\dfrac{1}{11\times12}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\\ =\dfrac{1}{4}-\dfrac{1}{12}\\ =\dfrac{3}{12}-\dfrac{1}{12}=\dfrac{2}{12}=\dfrac{1}{6}\)

\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{10}-\frac{1}{11}=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)

=1/2.3+1/3.4+1/4.5+......+1/10.11

=1-1/2+1/2-1/3+.....,+1/10-1/11

=1-1/11

=10/11

Tick