a) \(A=2^{2010}-2^{2009}-2^{2008}-...-2-1\)

\(A=2^{2010}\left(2^{2009}+2^{2008}+...+2+1\right)\)

Đặt \(\text{A = 1 + 2 + . . . + 2^{2008} + 2^{2009}}\)

\(\text{⇒ 2 A = 2 + 2 2 + . . + 2^{2010}}\)

⇒ \(A=2^{2010}-1\)

⇒ \(A=2^{2010}-\left(2^{2010}-1\right)\)

⇒ \(A=1\)

b) \(B=2072\)

c) \(\dfrac{4949}{19800}\)

Xin lỗi mình không có nhiều thời gian để giải thích trên đây á nên tạm gửi ảnh mình tạo nhé . Học tốt !

câu b bài 2:

\(\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\)

\(=\dfrac{1}{5}\)

câu a bài 2:

\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{10\cdot11\cdot12}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{12}\)

\(=1-\dfrac{1}{12}=\dfrac{11}{12}\)

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

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

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

Ta có: \(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\cdots+\frac{1}{98\cdot99\cdot100}\)

\(=\frac12\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\cdots+\frac{2}{98\cdot99\cdot100}\right)\)

\(=\frac12\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\cdots+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)

\(=\frac12\left(\frac12-\frac{1}{99\cdot100}\right)<\frac12\cdot\frac12=\frac14<2\)

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

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

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

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

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

\(S=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{2009.2010.2011}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{2009.2010}-\dfrac{1}{2010.2011}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2010.2011}\)

\(=\dfrac{1}{2}-\dfrac{1}{4042110}< \dfrac{1}{2}\)

\(\Rightarrow\) \(S< P\)

Vậy \(S< P\)