Giải:

\(\dfrac{5}{1.2}+\dfrac{5}{2.3}+\dfrac{5}{3.4}+...+\dfrac{5}{98.99}+\dfrac{5}{99.100}\) 

\(=5.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\) 

\(=5.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\) 

\(=5.\left(1-\dfrac{1}{100}\right)\) 

\(=5.\dfrac{99}{100}\) 

\(=\dfrac{99}{20}\) 

Chúc em học tốt!

Giải:

=5.(11.2+12.3+13.4+...+198.99+199.100)=5.(11.2+12.3+13.4+...+198.99+199.100) 

=5.(1−1100)=5.(1−1100) 

=9920=9920 

\(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100.101}\)

\(S=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{99.100}-\dfrac{1}{100.101}\right)\)

\(S=\dfrac{1}{4}-\dfrac{1}{2.100.101}\)

Ta có  1/1.2-1/2.3=2/1.2.3;1/2.3-1/3.4=2/2.3.4 .....1/98.99-1/99.100=2/98.99.100                                                                                               2A=2/1.2.3+2/2.3.4+....+2/98.99.100 = 1/1.2-1/2.3+1/2.3-1/3.4+...+1/98.99-1/99.100 = 1/2-1/99.100 = 4949/9900                                           A =4949/19800                                                                                                     

dễ ợt tự làm đê

máy tính đã chứng minh nhá

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{1}{k}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Leftrightarrow\frac{1}{2}=\frac{1}{k}\Rightarrow k=2\)

k=2

chuan 100%

k=2 do

\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{n+2-n}{n\left(n+1\right)\left(n+2\right)}=\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{n\left(n+2\right)}\)

\(\Rightarrow\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

\(=\frac{1}{1.2}-\frac{1}{99.100}\)

\(\Rightarrow\frac{1}{1.2.3}+...+\frac{1}{98.99.100}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Rightarrow k=2\)

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)

\(=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{100-98}{98.99.100}\right)=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)=\frac{1}{2}.\frac{4949}{9900}=\frac{4949}{18000}\)

=1-/2-1/3+1/2-1/3-1/4+1/5-1/6-1/7+1/6-1/7-1/8-.........-1/98-1/99-1/100

=1-1/100

=99/100

= 1 - 1/2 - 1/3 - 1/4 - ... - 1/98 - 1/99 - 1/100

= 1 - 1/100

= 99/100

Ta đặt

  \(A=\dfrac{7}{1\times2}+\dfrac{7}{2\times3}+...+\dfrac{7}{99\times100}\)

\(\dfrac{1}{7}\times A=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+....+\dfrac{1}{99\times100}\)

\(\dfrac{1}{7}\times A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\dfrac{1}{7}\times A=1-\dfrac{1}{100}\)

\(\dfrac{1}{7}\times A=\dfrac{99}{100}\)

\(A=\dfrac{99}{100}\div\dfrac{1}{7}\)

\(A=\dfrac{693}{100}\)

= 7.(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100)

= 7.(1 - 1/100)

= 7 . 99/100

= 693/100

\(A=7\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}=\)

\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{100-99}{99.100}=\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

\(\Rightarrow A=7x\dfrac{99}{100}=6,93\)

a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)

\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=5\left(1-\dfrac{1}{100}\right)\)

\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)

b, \(C=1.2.3+2.3.4+...+8.9.10\)

\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)

\(C=\dfrac{8.9.10.11}{4}=1980.\)

c, https://hoc24.vn/hoi-dap/question/384591.html

Câu này bạn vào đây mình đã giải câu tương tự nhé.

\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)

\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)

\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)

\(\Leftrightarrow A=\dfrac{99}{20}\)