c: \(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{49\cdot50}\)

\(=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{49}-\frac{1}{50}\)

\(=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{49}+\frac{1}{50}-2\left(\frac12+\frac14+\cdots+\frac{1}{50}\right)\)

\(=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{50}-1-\frac12-\cdots-\frac{1}{25}\)

\(=\frac{1}{26}+\frac{1}{27}+\cdots+\frac{1}{50}\)

giúp em câu a b nx dc hem tại khó quá em chx học kiểu chấm than ở mẫu số

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2023\cdot2024}\)

\(=\frac11-\frac12+\frac12-\frac13+\cdots+\frac{1}{2023}-\frac{1}{2024}\)

\(=\frac11-\frac{1}{2024}=\frac{2023}{2024}\)

A=1⋅21​+2⋅31​+3⋅41​+⋯+2023⋅20241​

\(= \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \hdots + \frac{1}{2023} - \frac{1}{2024}\)

\(= \frac{1}{1} - \frac{1}{2024} = \frac{2023}{2024}\)

Câu 1:

c: \(\frac19+\frac28+\frac37+\cdots+\frac91\)

\(=\left(\frac19+1\right)+\left(\frac28+1\right)+\cdots+\left(\frac82+1\right)+1\)

\(=\frac{10}{2}+\frac{10}{3}+\cdots+\frac{10}{10}=10\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)\)

Ta có: \(\left(\frac12+\frac13+\frac14+\cdots+\frac{1}{10}\right)\cdot x=\frac19+\frac28+\frac37+\cdots+\frac91\)

=>\(x\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)=10\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)\)

=>x=10

Câu 2:

d: \(\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+\cdots+\frac{1}{2021\cdot2022\cdot2023\cdot2024}\)

\(=\frac13\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+\cdots+\frac{1}{2021\cdot2022\cdot2023}-\frac{1}{2022\cdot2023\cdot2024}\right)\)

\(=\frac13\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2022\cdot2023\cdot2024}\right)\)

Hẹ hẹ

* Chứng minh \(\frac16

Ta có: \(F=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\cdots+\frac{1}{100^2}\)

\(F=\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\cdots+\frac{1}{100\cdot100}\)

\(\Rightarrow F<\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\cdots+\frac{1}{99\cdot100}\)

\(\) \(\Rightarrow F<\frac14-\frac15+\frac15-\frac16+\frac16-\frac17+\cdots+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow F<\frac14-\frac{1}{100}\)

\(\Rightarrow F<\frac{12}{25}\)

Mà \(\frac16=\frac{12}{72}<\frac{12}{25}\)

\(\Rightarrow\frac16 (1)

* Chứng minh \(F<\frac14\)

Lại có: \(\) \(F=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\cdots+\frac{1}{100^2}\)

\(F=\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\cdots+\frac{1}{100\cdot100}\)

\(\Rightarrow F>\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\cdots+\frac{1}{100\cdot101}\)

\(\Rightarrow F>\frac15-\frac16+\frac16-\frac17+\frac17-\frac18+\cdots+\frac{1}{100}-\frac{1}{101}\)

\(\Rightarrow F=\frac15-\frac{1}{101}\)

\(\Rightarrow F>\frac{96}{505}\)

Mà \(\frac14=\frac{96}{384}<\frac{96}{505}\)

\(\Rightarrow F<\frac14\) (2)

Từ (1) và (2) suy ra: \(\frac16

Vậy \(\frac16

Ta có: \(\frac{1}{1+\sqrt2}+\frac{1}{\sqrt2+\sqrt3}+\cdots+\frac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\frac{-1+\sqrt2}{\left(\sqrt2+1\right)\left(\sqrt2-1\right)}+\frac{-\sqrt2+\sqrt3}{\left(\sqrt3-\sqrt2\right)\left(\sqrt3+\sqrt2\right)}+\cdots+\frac{-\sqrt{99}+\sqrt{100}}{\left(\sqrt{100}+\sqrt{99}\right)\left(\sqrt{100}-\sqrt{99}\right)}\)

\(=-1+\sqrt2-\sqrt2+\sqrt3-\cdots-\sqrt{99}+\sqrt{100}\)

\(=-1+\sqrt{100}\)

=-1+10

=9

a: \(A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{99\cdot101}\)

\(=1-\frac13+\frac13-\frac15+\cdots+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}\)

b: \(B=\frac12-\left(\frac{1}{5\cdot11}+\frac{1}{11\cdot17}+\frac{1}{17\cdot23}+\frac{1}{23\cdot29}+\frac{1}{29\cdot35}\right)\)

\(=\frac12-\frac16\left(\frac{6}{5\cdot11}+\frac{6}{11\cdot17}+\frac{6}{17\cdot23}+\frac{6}{23\cdot29}+\frac{6}{29\cdot35}\right)\)

\(=\frac12-\frac16\left(\frac15-\frac{1}{11}+\frac{1}{11}-\frac{1}{17}+\cdots+\frac{1}{29}-\frac{1}{35}\right)\)

\(=\frac12-\frac16\left(\frac15-\frac{1}{35}\right)=\frac12-\frac16\cdot\frac{6}{35}=\frac12-\frac{1}{35}=\frac{33}{70}\)

\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+99\right)\left(x+100\right)}=\frac{k}{x\left(x+100\right)}\)

\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+99}-\frac{1}{x+100}=\frac{k}{x\left(x+100\right)}\)

\(\frac{1}{x}-\frac{1}{x+100}=\frac{k}{x\left(x+100\right)}\)

\(\frac{x+100}{x\left(x+100\right)}-\frac{x}{x\left(x+100\right)}=\frac{k}{x\left(x+100\right)}\)

k = 100

k=100

=1

Đặt \(A=\frac{1}{99}-\frac{1}{99\cdot98}-\frac{1}{98\cdot97}-\frac{1}{97\cdot96}-\cdots-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)

\(A=\frac{1}{99}-\left(\frac{1}{98}-\frac{1}{99}\right)-\left(\frac{1}{97}-\frac{1}{98}\right)-\left(\frac{1}{96}-\frac{1}{97}\right)-\cdots-\left(\frac12-\frac13\right)-\left(\frac11-\frac12\right)\)

\(A=\frac{1}{99}-\frac{1}{98}+\frac{1}{99}-\frac{1}{97}+\frac{1}{98}-\frac{1}{96}+\frac{1}{97}-\cdots-\frac12+\frac13-1+\frac12\)

\(A=\left(\frac{1}{99}-\frac{1}{99}\right)+\left(\frac{1}{98}-\frac{1}{98}\right)+\left(\frac{1}{97}-\frac{1}{97}\right)+\left(\frac{1}{96}-\frac{1}{96}\right)+\cdots+\left(\frac12-\frac12\right)-1\)

\(A=0+0+0+0+\cdots+0+\left(-1\right)\)

\(A=-1\)

Vậy A = -1

=-2. \(\left(\frac{-3}{2}\right).\left(\frac{-4}{3}\right).....\left(\frac{-2010}{2009}\right).\left(\frac{-2011}{2010}\right)\)

=\(\frac{\left(-2\right).\left(-3\right).\left(-4\right).....\left(-2010\right).\left(-2011\right)}{1.2.3.....2009.2010}\)

=\(\frac{\left(-1\right).\left(-1\right).\left(-1\right).....\left(-1\right).\left(-1\right).\left(-2011\right)}{1.1.1.....1.1}\)

=\(\frac{\left(-1\right)^{2009}.\left(-2011\right)}{1}\)

=\(\frac{\left(-1\right).\left(-2011\right)}{1}=\frac{2011}{1}=2011\)