Câu 2:
a) Đặt (12n + 1, 30n + 2) = d
\(\Rightarrow\) \(\left\{{}\begin{matrix}12n+1⋮d\\30n+2⋮d\end{matrix}\right.\) \(\Rightarrow\) \(\left\{{}\begin{matrix}5\left(12n+1\right)⋮d\\2\left(30n+2\right)⋮d\end{matrix}\right.\) \(\Rightarrow\) \(\left\{{}\begin{matrix}60n+5⋮d\\60n+4⋮d\end{matrix}\right.\)
\(\Rightarrow\) (60n + 5) - (60n + 4) \(⋮\) d
\(\Rightarrow\) 60n + 5 - 60n - 4 \(⋮\) d
\(\Rightarrow\) 1 \(⋮\) d
\(\Rightarrow\) \(d\in\left\{1;-1\right\}\)
\(\Rightarrow\) (12n + 1, 30n + 2) = \(\left\{1;-1\right\}\)
Vậy phân số \(\dfrac{12n+1}{30n+2}\) là phân số tối giản.
b) Ta thấy:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4};....;\dfrac{1}{100^2}=\dfrac{1}{100.100}< \dfrac{1}{99.100}\)
\(\Rightarrow\)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(\Rightarrow\) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{100^2}\)< \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow\) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{100^2}\) < \(1-\dfrac{1}{100}\) < 1
Vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{100^2}\) < 1