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

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

\(=\dfrac{1}{1.2}-\dfrac{1}{99.100}=\dfrac{50.99-1}{100.99}=\dfrac{4949}{9900}\)

`A=1/[1.2.3]+1/[2.3.4]+....+1/[98.99.100]`

`A=1/2.(2/[1.2.3]+2/[2.3.4]+....+2/[98.99.100])`

`A=1/2.(1/[1.2]-1/[2.3]+1/[2.3]-1/[3.4]+....+1/[98.99]-1/[99.100])`

`A=1/2.(1/[1.2]-1/[99.100])`

`A=1/2.(1/2-1/9900)`

`A=1/2.(4950/9900-1/9900)`

`A=1/2 . 4949/9900`

`A=4949/19800`

\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)

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

\(C=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

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

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

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 ~

* 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...

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

Lời giải:

Đặt biểu thức trên là $A$.
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{37.38.39}\)

\(=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)

\(=\frac{1}{1.2}-\frac{1}{38.39}=\frac{370}{741}\)

\(\Rightarrow A=\frac{185}{741}\)

\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2004.2005.2006}\)

\(=2.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)+2.\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+...+2.\left(\dfrac{1}{2004.2005}-\dfrac{1}{2005.2006}\right)\)

\(=2.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2004.2005}-\dfrac{1}{2005.2006}\right)\)

\(=2.\left(\dfrac{1}{1.2}-\dfrac{1}{2005.2006}\right)\)

\(=1-\dfrac{2}{2005.2006}\)

\(=\dfrac{2011014}{2011015}\).

Ta có:

\(M=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2004.2005.2006}\)

\(M=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{2004.2005.2006}\right)\)

\(M=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2004.2005}-\dfrac{1}{2005.2006}\right)\)

\(M=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2005.2006}\right)\)