a) Giải:
Ta có: \(4n-5=4\left(n-3\right)+7\)
Để \(\left(4n-5\right)⋮\left(n-3\right)\Leftrightarrow7⋮n-3\)
\(\Rightarrow n-3\inƯ\left(7\right)\)
Mà \(Ư\left(7\right)\in\left\{\pm1;\pm7\right\}\)
Nên ta có bảng sau:
| \(n-3\) | \(n\) |
| \(1\) | \(4\) |
| \(-1\) | \(2\) |
| \(-7\) | \(-4\) |
| \(7\) | \(10\) |
Vậy \(n=\left\{2;4;-4;10\right\}\)
b) Ta có:
\(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)
\(=\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)
Nhận xét:
\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{4}\)
\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}=\dfrac{1}{20}\)
\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)
Vậy \(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\) \(< \dfrac{1}{2}\) (Đpcm)
