a) đặt

 \(S=1+\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\\ 2S=2+\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\\ 2S=2+\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ 2S=2+1-\dfrac{1}{101}\\ 2S=\dfrac{302}{101}\\ S=\dfrac{151}{101}\)

b)

đặt

\(S=\dfrac{1}{2}+\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{98\cdot101}\\ 3S=\dfrac{3}{2}+\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{98\cdot101}\\ 3S=\dfrac{3}{2}+\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{98}-\dfrac{1}{101}\\ 3S=\dfrac{3}{2}+\dfrac{1}{2}-\dfrac{1}{101}\\ 3S=\dfrac{201}{101}\\ S=\dfrac{67}{101}\)

\(2A-1=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(2A-1=1-\dfrac{1}{101}=\dfrac{100}{101}\)

\(2A=\dfrac{201}{101}\Rightarrow A=\dfrac{201}{202}\)

\(\dfrac{n}{n-5}=\dfrac{n-5+5}{n-5}=\dfrac{n-5}{n-5}+\dfrac{5}{n-5}=1+\dfrac{5}{n-5}\)

Để \(\dfrac{n}{n-5}\in Z\) thì \(\dfrac{5}{n-5}\in Z\)

\(\Rightarrow n-5\in\text{Ư}_{\left(5\right)}=\left\{-5;-1;1;5\right\}\)

\(\Rightarrow n\in\left\{0;4;6;10\right\}\)

Do \(n\in N\) nên tất cả các giá trị đều nhận

Vậy ...

`(x-3/8) +1/4= 7/8`

`=> x-3/8 = 7/8-1/4`

`=> x-3/8 = 7/8-2/8`

`=> x-3/8 =5/8`

`=>x= 5/8 +3/8`

`=>x= 8/8=1`

khó thế =))

\(S=\dfrac{2^2}{3x5}+\dfrac{2^2}{5x7}+\dfrac{2^2}{7x9}+...+\dfrac{2^2}{97x99}\)

\(\dfrac{S}{2}=\dfrac{2}{3x5}+\dfrac{2}{5x7}+\dfrac{2}{7x9}+...+\dfrac{2}{97x99}\)

\(\dfrac{S}{2}=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}...+\dfrac{1}{97}-\dfrac{1}{99}=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)

S=\(\dfrac{64}{99}\)

Ta có : \(A\text{=}\dfrac{2023^{2023}}{2023^{2024}}\text{=}\dfrac{1}{2023}\)

và \(B\text{=}\dfrac{2023^{2022}}{2023^{2023}}\text{=}\dfrac{1}{2023}\)

\(\Rightarrow A\text{=}B\)

Ta có :

A=\(\dfrac{2023^{2023}}{2023^{2024}}\)=\(\dfrac{2023^{2022}.2023}{2023^{2023}.2023}\)=\(\dfrac{2023^{2022}}{2023^{2023}}\)

Mà B=\(\dfrac{2023^{2023}}{2023^{2024}}\)

Vậy A=B

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