Ta có :

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

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

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

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

\(S=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)\)

\(S=\dfrac{1}{2}.\dfrac{4949}{9900}\)

\(S=\dfrac{4949}{19800}\)

~ Chúc bn học tốt ~

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

* Áp dụng công thức: \(\dfrac{k}{n.\left(n+k\right)}\)=\(\dfrac{1}{n}-\dfrac{1}{n+k}\)

ta có : \(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-....+\dfrac{1}{98.99}-\dfrac{1}{99.100}\)

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

E= ........(tính ra)

E=4949/9900

Giải:

\(E=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}.\)

Áp dung tính chất:

\(\dfrac{2m}{b\left(b+1\right)\left(b+2\right)}=\dfrac{1}{b\left(b+1\right)}-\dfrac{1}{\left(b+m\right)\left(b+2\right)}\), ta có:

\(2E=2\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\right).\)

\(2E=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{98.99.100}.\)

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

\(2E=\dfrac{1}{1.2}+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{98.99}-\dfrac{1}{98.99}\right)-\dfrac{1}{99.100}.\)

\(2E=\dfrac{1}{1.2}+0+0+...+0-\dfrac{1}{99.100}.\)

\(2E=\dfrac{1}{1.2}-\dfrac{1}{99.100}.\)

\(2E=\dfrac{1}{2}-\dfrac{1}{9900}.\)

\(2E=\dfrac{4950}{9900}-\dfrac{1}{9900}.\)

\(2E=\dfrac{4949}{9900}.\)

\(\Rightarrow E=\dfrac{4949}{9900}:2.\)

\(\Rightarrow E=\dfrac{4949}{9900}.\dfrac{1}{2}=\dfrac{4949}{19800}.\)

Vậy \(E=\dfrac{4949}{19800}.\)

~ Học tốt!!! ~

* Chứng tỏ

Ta có :\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)

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

= \(\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{98.99.100}\right)\)

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

= \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}+0+0+...+0+\dfrac{-1}{99.100}\right)\)

= \(\dfrac{1}{2}.\left(\dfrac{1}{2}+\dfrac{-1}{9900}\right)\)

= \(\dfrac{1}{2}.\left(\dfrac{4850}{9900}+\dfrac{-1}{9900}\right)\)

= \(\dfrac{1}{2}.\dfrac{4849}{9900}\)

= \(\dfrac{4849}{19800}\)

* So sánh

\(\dfrac{4950}{19800}\) và \(\dfrac{1}{4}\)

\(\dfrac{1}{4}=\dfrac{4950}{19800}\)

Vì \(\dfrac{4950}{19800}=\dfrac{4950}{19800}\)

=> Tổng trên bằng với\(\dfrac{1}{4}\)

mình ko chắc là đúng

a) Ta có: \(3xy+x-3y=6\)

\(\Rightarrow x\left(3y+1\right)-3y=6\)

\(\Rightarrow x\left(3y+1\right)-\left(3y+1\right)=5\)

\(\Rightarrow\left(x-1\right)\left(3y+1\right)=5\)

Ta có bảng sau:

....

b) Ta có: \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)

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

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

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

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

\(=\frac{1}{2}.\frac{4949}{9900}\)

\(=\frac{4949}{19800}\)

\(\Rightarrow\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}=\frac{4949}{19800}\left(đpcm\right)\)

Vậy...

\(S=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{98\cdot99\cdot100}\\ =\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{98\cdot99\cdot100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}-\dfrac{1}{99\cdot100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)\\ =\dfrac{1}{2}\cdot\dfrac{4949}{9900}\\ =\dfrac{4949}{19800}\)

\(A=\left(\dfrac{-1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{4}{35}+\dfrac{2}{7}\right)+\dfrac{1}{127}\)

\(A=\left(\dfrac{-9-2-7}{18}\right)+\left(\dfrac{21+4+10}{35}\right)+\dfrac{1}{127}\)

\(A=-1+1+\dfrac{1}{127}\)

\(A=\dfrac{1}{127}\)

\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)

\(\dfrac{1}{4}B=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.4}+\dfrac{1}{3.4.5.4}+...+\dfrac{1}{98.99.100.4}\)

\(\dfrac{1}{4}B=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.\left(5-1\right)}+\dfrac{1}{3.4.5.\left(6-2\right)}+...+\dfrac{1}{98.99.100.\left(101-97\right)}\)

\(\dfrac{1}{4}B=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5-1.2.3.4}+\dfrac{1}{3.4.5.6-2.3.4.5}+...+\dfrac{1}{98.99.100.101-97.98.99.100}\)

\(\dfrac{1}{4}B=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}-\dfrac{1}{1.2.3.4}+\dfrac{1}{3.4.5.6}-\dfrac{1}{2.3.4.5}+...+\dfrac{1}{98.99.100.101}-\dfrac{1}{97.98.99.100}\)

\(\dfrac{1}{4}B=\dfrac{1}{98.99.100.101}\)

\(B=\dfrac{1}{98.99.100.101}.4=\dfrac{1}{98.99.25.101}\)

tick cho mk nha

bài tự làm 100%

co gì chưa đc thì coi lại nha

\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{37.38.39}\)

\(A=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{37.38.39}\right)\)

\(A=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{37.38}-\dfrac{1}{38.39}\right)\)

\(A=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{38.39}\right)\)

\(A=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{1482}\right)\)

\(A=\dfrac{1}{2}.\dfrac{370}{741}=\dfrac{185}{741}\)