Đặt \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)

Ta có: \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)

\(\Leftrightarrow2A=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\dfrac{2}{3\cdot4\cdot5}+...+\dfrac{2}{98\cdot99\cdot100}\)

\(\Leftrightarrow2A=-\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}-\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}-\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}-\dfrac{1}{4\cdot5}+...-\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)

\(\Leftrightarrow2A=-\dfrac{1}{2}+\dfrac{1}{99\cdot100}\)

\(\Leftrightarrow2A=\dfrac{-1}{2}+\dfrac{1}{9900}\)

\(\Leftrightarrow2A=\dfrac{-4950}{9900}+\dfrac{1}{9900}=\dfrac{-4949}{9900}\)

hay \(A=\dfrac{-4949}{19800}\)

 \(\text{Charlotte :'(}\)

Giải phương trình.

 \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{x\left(x+1\right)\left(x+2\right)}=\frac{637}{2550}\)   \(\left(\text{*}\right)\)  

\(ĐKXĐ:\)  \(x\ne0;\)  \(x\ne-1;\)   và  \(x\ne-2\)

Ta có:

\(\frac{1}{1.2.3}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)\)

\(\frac{1}{2.3.4}=\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)\)

\(\frac{1}{3.4.5}=\frac{1}{2}\left(\frac{1}{3.4}-\frac{1}{4.5}\right)\)

\(.....................\)

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

Khi đó, phương trình   \(\left(\text{*}\right)\)   tương đương với  

 \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}-\frac{1}{\left(x+1\right)\left(x+2\right)}\right)=\frac{637}{2550}\)

\(\Leftrightarrow\)  \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{\left(x+1\right)\left(x+2\right)}\right)=\frac{637}{2550}\)

\(\Leftrightarrow\)  \(\frac{1}{4}-\frac{1}{2\left(x+1\right)\left(x+2\right)}=\frac{637}{2550}\)

\(\Leftrightarrow\)  \(\frac{1}{2\left(x+1\right)\left(x+2\right)}=\frac{1}{5100}\)

\(\Rightarrow\)   \(2\left(x+1\right)\left(x+2\right)=5100\)

\(\Leftrightarrow\)  \(\left(x+1\right)\left(x+2\right)=2550\)

\(\Leftrightarrow\)  \(^{x_1=-52}_{x_2=49}\)  (t/m điều kiện xác định)

Vậy,  tập nghiệm của pt  \(\left(\text{*}\right)\)  là  \(S=\left\{-52;49\right\}\)

=1/2-1/3-1/4+1/3-1/4-1/5+1/5-1/6-1/7+...+1/35-1/36-1/37

giao hoán, kết hợp là ra nha

Đặt \(A=\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{30\times31\times32}\)

\(2A=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+\frac{2}{3\times4\times5}+...+\frac{2}{30\times31\times32}\)

\(=\left(\frac{1}{1\times2}-\frac{1}{2\times3}\right)+\left(\frac{1}{2\times3}-\frac{1}{3\times4}\right)+\left(\frac{1}{3\times4}-\frac{1}{4\times5}\right)+...+\left(\frac{1}{30\times31}-\frac{1}{31\times32}\right)\)

\(=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{3\times4}-\frac{1}{4\times5}+...+\frac{1}{30\times31}-\frac{1}{31\times32}\)

\(=\frac{1}{1\times2}-\frac{1}{31\times32}\)

\(=\frac{1}{2}-\frac{1}{992}\)

\(=\frac{495}{992}\)

\(\Rightarrow A=\frac{495}{992}\div2=\frac{495}{1984}\)

\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{30\times31\times32}\)

\(=\frac{1}{2}\times\left(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{30\times31\times32}\right)\)

\(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{3\times4}-\frac{1}{4\times5}+...+\frac{1}{30\times31}-\frac{1}{31\times32}\right)\)

\(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{31\times32}\right)\)

\(=\frac{1}{2}\times\frac{990}{1984}\)

\(=\frac{990}{3968}=\frac{495}{1984}\)

Tham khảo bài mik tại link này nhé :

https://olm.vn/hoi-dap/detail/197648362685.html

S = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2013.2014.2015}\)

S = \(\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+....+\frac{2015-2013}{2013.2014.2015}\right)\)

S = \(\frac{1}{2}.\left(\frac{3}{1.2.3}-\frac{1}{1.2.3}+\frac{4}{2.3.4}-\frac{2}{2.3.4}+...+\frac{2015}{2013.2014.2015}-\frac{2013}{2013.2014.2015}\right)\)

S = \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2013.2014}-\frac{1}{2014.2015}\right)\)

S = \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2014.2015}\right)\)

S = \(\frac{1}{2}.\frac{2029104}{4058210}\)

S = \(\frac{1014552}{4058210}\)

`A=1/[1xx2xx3]+1/[2xx3xx4]+1/[3xx4xx5]+....+1/[98xx99xx100]`

`A=1/2xx(2/[1xx2xx3]+2/[2xx3xx4]+2/[3xx4xx5]+....+2/[98xx99xx100])`

`A=1/2xx(1/[1xx2]-1/[2xx3]+1/[2xx3]-1/[3xx4]+1/[3xx4]-1/[4xx5]+....+1/[98xx99]-1/[99xx100])`

`A=1/2xx(1/[1xx2]-1/[99xx100])`

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

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

`A=1/2xx4949/9900`

`A=4949/19800`

\(A=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}\)

\(A=\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right):2\)

\(A=\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{12}-\dfrac{1}{20}+...+\dfrac{1}{9702}-\dfrac{1}{990}\right):2\)

\(A=\left(\dfrac{1}{2}-\dfrac{1}{990}\right):2\)

\(A=\dfrac{4949}{9900}:2\)

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

\(=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\)

=1/2-1/380

=190/380-1/380

=189/380

Gọi biểu thức trên là S. Ta có :

\(S=\dfrac{1}{1\times2\times3}+\dfrac{1}{2\times3\times4}+\dfrac{1}{3\times4\times5}+...+\dfrac{1}{18\times19\times20}\)

\(=\dfrac{1}{2}\times\left(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+...+\dfrac{2}{18\times19\times20}\right)\)

Trước tiên, ta áp dụng : \(\dfrac{2}{a\left(a+1\right)\left(a+2\right)}=\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{\left(a+1\right)\left(a+2\right)}\)

Ta sẽ có : 

\(S=\dfrac{1}{2}\times\left(\dfrac{1}{1\times2}-\dfrac{1}{2\times3}+\dfrac{1}{2\times3}-\dfrac{1}{3\times4}+\dfrac{1}{3\times4}-\dfrac{1}{4\times5}+...+\dfrac{1}{18\times19}-\dfrac{1}{19\times20}\right)\)

\(=\dfrac{1}{2}\times\left(\dfrac{1}{1\times2}-\dfrac{1}{19\times20}\right)\)

\(=\dfrac{1}{2}\times\dfrac{1}{1\times2}-\dfrac{1}{2}\times\dfrac{1}{19\times20}\)

\(=\dfrac{1}{4}-\dfrac{1}{760}=\dfrac{189}{760}\)

`=1/2(1/1×2 - 1/2×3 + 1/2×3 - 1/3×4 + 1/3×4 - 1/4×5 + ... + 1/18×19 - 1/19×20)`
`=1/2(1/2 - 1/19×20)`
`=1/2×189/380 `
`=189/760`

A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100

A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3

A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)

A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.

A x 3 = 99x100x101

A = 99x100x101 : 3

A = 333300 

Ta có:

\(A=1.2+2.3+3.4+...+99.100\)

\(\Rightarrow3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)

\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)

\(\Leftrightarrow3A=99.100.101\Leftrightarrow A=\frac{99.100.101}{3}=333300\)

\(B=1.2.3+2.3.4+4.5.6+...+98.99.100\)

\(\Rightarrow4B=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+4.5.6.\left(7-3\right)+...+98.99.100.\left(101-97\right)\)

\(\Rightarrow4B=1.2.3.4+2.3.4.5-1.2.3.4+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100\)

\(\Leftrightarrow4B=98.99.100.101\Leftrightarrow B=\frac{98.99.100.101}{4}=24497550\)

A= 1 x 2 + 2 x 3 + 3 x 4 +........+ 99 x 100

=> 2 + 6 + 12 +........+ 9900

=> 8 + 12 +.....+ 9900

=> 20 +....+ 9900

=> 20 + 20 + 30 +....+ 9900

=> 70 +....+ 9900

=> ( 9900 x 70 ) : 2

=> 693000 : 2

=> 346500

B = 1 x 2 x 3 + 2 x 3 x 4 +......+ 98 x 99 x100

=> ( 1 x 2 x 3) + ( 2 x 3 x 4 ) +....+ ( 98 x 99 x 100 )

= 6 + 24 +.......+ 970200

=> 28 + 120 +...+ 970200

=> ( 148 x 970200 ) : 2

=> 143589600 : 2

=> 71794900