Câu 1:Tìm tất cả số nguyên dương n sao cho tât cả số n+1,n+5,n+7,n+13,+n+17,n+25,n+37 đều là số nguyên tố
Câu 2:Cho A=\(1-\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
và B=\(\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}+\frac{1}{2019}\).Tính \(\left(A-B-1\right)^{2019}\)
1. \(n\in\left\{1;2;3;4;5;...\right\}\)
2. \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
\(\Rightarrow A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(\Rightarrow A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{1009}\)
\(\Rightarrow A=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}\)
Ta có :
\(\left(A-B-1\right)^{2019}=\left(\frac{1}{1010}+...+\frac{1}{2019}-\left(\frac{1}{1010}+...+\frac{1}{2019}\right)-1\right)^{2019}\)
\(=\left(-1\right)^{2019}=-1\)