\(S=\dfrac{4}{1.2.3}-\dfrac{1}{1.2.3}+\dfrac{6}{2.3.4}-\dfrac{1}{2.3.4}+...+\dfrac{4018}{2008.2009.2010}-\dfrac{1}{2008.2009.2010}\)

\(=\left(\dfrac{2}{1.3}+\dfrac{2}{2.4}+...+\dfrac{2}{2008.2010}\right)-\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2008.2009.2010}\right)\)

\(=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2007.2009}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2008.2010}\right)-\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{2008.2009.2010}\right)\)

\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)-\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}-\dfrac{1}{2009.2010}\right)\)

\(=\left(1-\dfrac{1}{2009}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)-\left(\dfrac{1}{1.2}-\dfrac{1}{2009.2010}\right)\)

\(=1-\dfrac{1}{2009}-\dfrac{1}{2010}+\dfrac{1}{2009.2010}\)

\(=\dfrac{1}{2010}\left(\dfrac{1}{2009}-1\right)-\left(\dfrac{1}{2009}-1\right)\)

\(=\left(\dfrac{1}{2010}-1\right)\left(\dfrac{1}{2009}-1\right)=\dfrac{2009}{2010}.\dfrac{2008}{2009}=\dfrac{1004}{1005}\)

Gọi biểu thức là \(A\). Ta có :

\(A=\dfrac{3}{1.2.3}+\dfrac{5}{2.3.4}+\dfrac{7}{3.4.5}+...+\dfrac{2017}{1008.1009.1010}\)

\(A=\left(\dfrac{1.2}{1.2.3}+\dfrac{2.2}{2.3.4}+\dfrac{3.2}{3.4.5}+...+\dfrac{1008.2}{1008.1009.1010}\right)+\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{1008.1009.1010}\right)\)\(A=\left(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{1009.1010}\right)+\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{1008.1009}-\dfrac{1}{1009.1010}\right)\)

\(A=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{1009}-\dfrac{1}{1010}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{1009.1010}\right)\)

\(A< 2.\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{1}{2}=1+\dfrac{1}{4}=\dfrac{5}{4}\)

Ta có: \(\frac{3n+2}{n\left(n+1\right)\left(n+2\right)}\)

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

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

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

Do đó, ta có: \(\frac{5}{1\cdot2\cdot3}=\frac{3\cdot1+2}{1\cdot2\cdot3}=\frac11+\frac{1}{1+1}-\frac{2}{1+2}=1+\frac12-\frac23\)

\(\frac{8}{2\cdot3\cdot4}=\frac{3\cdot2+2}{2\cdot3\cdot4}=\frac12+\frac13-\frac24\)

...

Do đó, ta có: \(S=1+\frac12-\frac23+\frac12+\frac13-\frac24+\frac13+\frac14-\frac25+\ldots+\frac{1}{n}+\frac{1}{n+1}-\frac{2}{n+2}\)

\(=1+\left(\frac12+\frac12\right)+\left(-\frac23+\frac13+\frac13\right)+\left(-\frac24+\frac14+\frac14\right)+\cdots+\left(-\frac{2}{n}+\frac{1}{n}+\frac{1}{n}\right)-\frac{2}{n+1}+\frac{1}{n+1}-\frac{2}{n+2}\)

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

=>\(S_{2022}=\frac{5}{1\cdot2\cdot3}+\frac{8}{2\cdot3\cdot4}+\cdots+\frac{6068}{2022\cdot2023\cdot2024}<2\)

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

=1/2-1/380

=179/380

b: \(=\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{21\cdot23}-\dfrac{1}{23\cdot25}\)

\(=\dfrac{1}{3}-\dfrac{1}{575}=\dfrac{572}{1725}\)

c: \(=1+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}-\dfrac{1}{20}-\dfrac{1}{21}\)

=1-1/21

=20/21

d: \(=\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{121}\right)\)

\(=\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{10}{11}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{12}{11}\)

\(=\dfrac{2}{11}\cdot\dfrac{12}{2}=\dfrac{12}{11}\)

Đặt S=A

Ta có: \(\frac{3n+2}{n\left(n+1\right)\left(n+2\right)}\)

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

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

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

Do đó, ta có: \(\frac{5}{1\cdot2\cdot3}=\frac{3\cdot1+2}{1\cdot2\cdot3}=\frac11+\frac{1}{1+1}-\frac{2}{1+2}=1+\frac12-\frac23\)

\(\frac{8}{2\cdot3\cdot4}=\frac{3\cdot2+2}{2\cdot3\cdot4}=\frac12+\frac13-\frac24\)

...

Do đó, ta có: \(S=1+\frac12-\frac23+\frac12+\frac13-\frac24+\frac13+\frac14-\frac25+\ldots+\frac{1}{n}+\frac{1}{n+1}-\frac{2}{n+2}\)

\(=1+\left(\frac12+\frac12\right)+\left(-\frac23+\frac13+\frac13\right)+\left(-\frac24+\frac14+\frac14\right)+\cdots+\left(-\frac{2}{n}+\frac{1}{n}+\frac{1}{n}\right)-\frac{2}{n+1}+\frac{1}{n+1}-\frac{2}{n+2}\)

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

=>\(S_{2018}<2\)

Tìm y:

-y:1/2-5/2=4+1/2

-y:1/2 = 4+1/2+5/2

-y:1/2 = 7

-y = 7.2

y = -14

Vậy y = -14