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

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

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

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

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

\(\Rightarrow k=2\)

\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right).y=\frac{49}{100}\)

\(\Leftrightarrow\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\right).y=\frac{49}{100}\)

\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right).y=\frac{49}{100}\)

\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{99.100}\right).y=\frac{49}{100}\Leftrightarrow\left(\frac{99.50-1}{99.100}\right).y=\frac{49}{100}\)

\(\Leftrightarrow\left(\frac{99.50-1}{99}\right).y=49\Leftrightarrow\left(99.50-1\right).y=99.49\Rightarrow y=\frac{99.49}{99.50-1}\)

ảnh đại diện đẹp thế lấy ở đâu

Ta có : \(S=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{99.100}\right)\)

\(\Rightarrow S=2.\left(1-\frac{1}{100}\right)\)

\(=2.\frac{99}{100}=\frac{99}{50}\)

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

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

S= 2.\(\frac{99}{100}\)

S=\(\frac{99}{50}\)

\(S=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{98.99}+\frac{2}{99.100}\)

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

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

\(S=2.\left(1-\frac{1}{100}\right)\)

\(S=2.\frac{99}{100}\)

\(S=\frac{99}{50}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

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

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

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

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

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

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

ta có : 1/1,2 + 1/2,3 + 1/3,4 + ... + 1/98,99 + 1/99,100

=> : 1 - 1/2 + 1/2 - 1/3 + 1/2 - 1/4 +... 1/99 - 1/100

=> : 1- 1/100 = ?

=> : 99/100

A = 9(\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\))

A = 9(\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\))

A = 9(1 - \(\frac{1}{100}\))

A = 9.\(\frac{99}{100}\)=\(\frac{891}{100}\)=8,91

Vì \(\frac{1}{1.2}=\frac{1}{2}\)

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

Nên trong bài toán: \(\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\)

Mấy cái kia cũng vậy nên bạn yên tâm nha!!!!

A : 9 = 1/1.2 + 1/2.3 + 1/3.4 + ..... + 1/98.99 + 1/99.100

A : 9 = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ..... + 1/98 - 1/99 + 1/99 - 1/100

A : 9 = 1 - 1/100

A : 9 = 100/100 - 1/100

A : 9 = 99/100

A = 9 . 99/100

A = 891/100 = 8,91 = 8 91/100

\(E=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)

\(E=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\)

\(E=\dfrac{1}{1.2}-\dfrac{1}{99.100}=\dfrac{4949}{9900}\)

\(E=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)

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

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

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right)\)

...

E = \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)

2E = \(\dfrac{2}{1.2.3}+\dfrac{2}{3.4.5}+...+\dfrac{2}{98.99.100}\)

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

= \(\dfrac{1}{1.2}\) - \(\dfrac{1}{99.100}\)= \(\dfrac{1}{2}-\dfrac{1}{9900}\) = \(\dfrac{4949}{9900}\)