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

=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 =1x2x3 + 2x3x4 +3x4x5+....+ 2010 x2011 x 2012

4A =1x2x3x4 + 2x3x4x4 +3x4x5x4+....+ 2010 x2011 x 2012x4

4A =1x2x3x4 + 2x3x4x(5+1) +3x4x5x(6-2)+....+ 2010 x2011 x 2012x(2013-2009)

4A =1x2x3x4 + 2x3x4x5-1x2x3x4+3x4x5x6-2x3x4x5+....+ 2010 x2011 x 2012x2013-2009x2010x2011x2012

4A = 2010 x2011 x 2012x2013

A = \(\frac{2010\times2011\times2012\times2013}{4}\)

đặt S=1.2.3+2.3.4+....+18.19.20

4S=1.2.3.4+2.3.4.(5-1)+.......+18.19.20.(21-17)

4S=1.2.3.4-1.2.3+2.3.4.5-1.2.3.4+......+18.19.20.21-17.18.19.20

4S=....tự làm nha

bonking copy đó

..đề đến 2010,2011,2012 à...

nhục thật ^^

haizz công nhận trên đời còn có mấy loại.....ok mk thì hay lắm mà

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 \(A=\dfrac{-4949}{19800}\)

2A=\(\frac{2}{1.2.3}\)+\(\frac{2}{2.3.4}\)+...+\(\frac{2}{18.19.20}\)

=1/1.2-1/2.3+1/2.3-1/3.4+...+1/18.19-1/19.20

=1/2-1/19.20

A=1/4-1/19.20.2

vậy A<1/4

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)=\frac{1}{4}-\frac{1}{760}< \frac{1}{4}\)(ĐPCM)

\(\Leftrightarrow3x-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\right)=\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\right)\)

\(\Leftrightarrow3x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{2.3}+....+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(\Leftrightarrow3x-\left(1-\frac{1}{100}\right)=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)

\(\Leftrightarrow3x-\frac{99}{100}=\frac{1}{2}\cdot\frac{189}{380}\)

\(\Leftrightarrow3x-\frac{99}{100}=\frac{189}{760}\)

\(\Leftrightarrow3x=\frac{189}{760}+\frac{99}{100}=\frac{4707}{3800}\)

\(\Leftrightarrow x=\frac{1569}{3800}\)

\(\text{Vậy }x=\frac{1569}{3800}\)

Học sinh gương mẫu của lớp thầy Phú là đây

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