Question

[lưu trâm]

giúp mk với mn 

1) 

Tính: A  = \(\dfrac{2}{1.2}\) + \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + ...... + \(\dfrac{2}{2023.2024}\)

2)

a) Tìm số nguyên n sao cho: \(\dfrac{n}{n+2}\) + \(\dfrac{5}{n+2}\) là số nguyên

b) Tính tổng: S = \(\dfrac{1}{1.3}\) + \(\dfrac{1}{3.5}\) + \(\dfrac{1}{5.7}\) + .... + \(\dfrac{1}{2023.2025}\)

mk cảm ơn

Answer 1

[Sahara]

1)\(A=\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{2023.2024}\)
\(=2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2023.2024}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\right)\)
\(=2\left(1-\dfrac{1}{2024}\right)\)
\(=2\cdot\dfrac{2023}{2024}=\dfrac{2023}{1012}\)
2)
a/\(\dfrac{n}{n+2}+\dfrac{5}{n+2}=\dfrac{n+5}{n+2}=\dfrac{\left(n+2\right)+3}{n+2}=1+\dfrac{3}{n+2}\)
Để \(\dfrac{n}{n+2}+\dfrac{5}{n+2}\) là số nguyên thì \(n+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Ta có bảng sau:

\(n+2\)	\(-3\)	\(-1\)	\(1\)	\(3\)
\(n\)	\(-5\)	\(-3\)	\(-1\)	\(1\)

Vậy để \(\dfrac{n}{n+2}+\dfrac{5}{n+2}\) nguyên thì \(n\in\left\{-5;-3;-1;1\right\}\)
b/\(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2023.2025}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2023.2025}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2023}-\dfrac{1}{2025}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{2025}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2024}{2025}=\dfrac{1012}{2025}\)
 

Answer 2

[Coin Hunter]

1) \(A=\dfrac{2}{1.2}+\dfrac{2}{2.3}+...+\dfrac{2}{2023.2024}\)

   \(A=2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2023.2024}\right)\\ A=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\right)\\ A=2\left(1-\dfrac{1}{2024}\right)\\ A=2\cdot\dfrac{2023}{2024}=\dfrac{2023}{1012}\)