Bài 2:
a) \(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
\(\Rightarrow\dfrac{1}{3}\cdot\left[\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{x\left(x+3\right)}\right]=\dfrac{101}{1540}\)
\(\Rightarrow\dfrac{1}{3}\cdot\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)
\(\Rightarrow\dfrac{1}{3}\cdot\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)
\(\Rightarrow\dfrac{1}{3}\cdot\dfrac{x-2}{5\left(x+3\right)}=\dfrac{101}{1540}\)
\(\Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\)
\(\Rightarrow1540\left(x-2\right)=101\cdot15\left(x+3\right)\)
\(\Rightarrow1540x-3080=1515x+4545\)
\(\Rightarrow1540x-1515x=4545+3080\)
\(\Rightarrow25x=7625\)
\(\Rightarrow x=\dfrac{7625}{25}=305\)
b) Ta có:
\(10A=\dfrac{10^{2025}+10}{10^{2025}+1}=\dfrac{10^{2025}+1+9}{10^{2025}+1}=1+\dfrac{9}{10^{2025}+1}\)
\(10B=\dfrac{10^{2006}+10}{10^{2006}+1}=\dfrac{10^{2006}+1+9}{10^{2006}+1}=1+\dfrac{9}{10^{2006}+1}\)
Mà: \(10^{2025}+1>10^{2006}+1\)
\(\Rightarrow\dfrac{9}{10^{2025}+1}< \dfrac{9}{10^{2006}+1}\)
\(\Rightarrow1+\dfrac{9}{10^{2025}+1}< 1+\dfrac{9}{10^{2006}+1}\)
\(\Rightarrow10A< 10B\)
\(\Rightarrow A< B\)
c) Gọi d là ƯCLN(2n+7; n+2)
Ta có:
2n + 7 ⋮ d và n + 2 ⋮ d
⇒ 2n + 7 ⋮ d và 2(n + 2) ⋮ d
⇒ 2n + 7 ⋮ d và 2n + 4 ⋮ d
⇒ 2n + 7 - 2n - 4 ⋮ d
⇒ 3 ⋮ d
⇒ d ∈ {1; 3}
Để `(2n + 7)/(n+2)` là phân số tối giản thì:
`d ≠ 3`
⇒ n + 2 không chia hết cho 5
⇒ n + 2 ≠ 5k
⇒ n ≠ 5k - 2
