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
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/\(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2023.2025}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2023.2025}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2023}-\dfrac{1}{2025}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{2025}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2024}{2025}=\dfrac{1012}{2025}\)
1) \(A=\dfrac{2}{1.2}+\dfrac{2}{2.3}+...+\dfrac{2}{2023.2024}\)
\(A=2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2023.2024}\right)\\ A=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\right)\\ A=2\left(1-\dfrac{1}{2024}\right)\\ A=2\cdot\dfrac{2023}{2024}=\dfrac{2023}{1012}\)
2)
a) \(\dfrac{n}{n+2}+\dfrac{5}{n+2}là\text{ }số\text{ }nguyên\text{ }khi\text{ }\dfrac{n+5}{n+2}là\text{ }sốn\text{ }nguyên\)
\(\dfrac{n+5}{n+2}là\text{ }số\text{ }nguyên\text{ }khi\text{ }n+5⋮n+2\)
\(n+5⋮n+2\\ \Leftrightarrow n+2+3⋮n+2\\ \Rightarrow3⋮n+2\left(vì\text{ }n+2⋮n+2\right)\\ \Rightarrow n+2\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\\ \Rightarrow n\in\left\{-1;-3;1;-5\right\}\)
Vậy ...
b) \(S=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2023.2025}\right)\\ S=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2023}-\dfrac{1}{1-2025}\right)\\ S=\dfrac{1}{2}\left(1-\dfrac{1}{2025}\right)\\ S=\dfrac{1}{2}\cdot\dfrac{2024}{2025}=\dfrac{1012}{2025}\)
Vậy ...
