\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{110}\)

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+....+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)

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

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

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

\(\frac{1}{2}+\frac{1}{3}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}\)\(+\frac{1}{110}\) 

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...\) \(+\frac{1}{9\cdot10}\)\(+\frac{1}{10\cdot11}\)

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

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

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

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\)

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

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

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

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

= \(1-\frac{1}{11}=\frac{10}{11}\)

tôi bị sai sô \(\frac{1}{110}\)nha mọi  người thông cảm

3581/3960 nha bn

K mình nha

1/2+1/6+1/12+...+1/110

=1/1.2+1/2.3+1/3.4+...+1/10.11

=1-1/2+1/2-1/3+1/3-1/4+...+1/10-1/11

=1-1/11=10/11

1/2

a)     5/30+15/90+25/150+35/210+45/270

       =1/6+1/6+1/6+1/6+1/6

       =1/6 x 5

       =5/6

b)     1/2+1/6+1/12+1/20+....+1/56

        =1/1x2+1/2x3+1/3x4+1/4x5+.....1/7x8

        =1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.......-1/7+1/7-1/8

        =1/1-1/8

         =7/8

c)     mình chịu

thank you bn nhìu nha

chịu,khó quá,nhất là câu c ý

=10/11 đó

mk làm tắt dc ko

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)

=\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}=1-\frac{1}{11}=\frac{10}{11}\)

Chỉ cần viết ra là: \(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}=1-\frac{1}{11}=\frac{10}{11}\)

1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + 1/6x7 + 1/7x8 + 1/8x9 + 1/9x10 + 1/10x11

= 1 - 1/2 + 1/2 - 1/3 + 1/3 -1/4 + 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 - 1/11 = 10/11

Đúng cho mình nha

ta gọi \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\)là A

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(\Leftrightarrow1.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)

\(\Rightarrow A=1-\frac{1}{10}=\frac{9}{10}\)

ta gọi B là biểu thức thứ2

\(B=\frac{2.2}{3}\times\frac{3.3}{2.4}\times\frac{4.4}{3.5}\times...\times\frac{10.10}{9.11}\)

\(\Rightarrow\)2 x \(\frac{10}{11}\)\(=\frac{20}{11}\)

\(\Rightarrow\)\(x+\frac{9}{10}=\frac{20}{11}+\frac{9}{110}\)

\(\Rightarrow x=1\)

mk nghĩ vậy bạn ạ, mk mong nó đúng

A = \(\frac{-79}{90}\)

B = \(\frac{8}{9}\)

cách giải sao chỉ mình với