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Bạn có thể viết lại đề theo phân số như thế này được không \(\frac{7}{12}\)bạn viết thế mk ko hiểu
Bn viết lại đề nhanh mk làm cho
Chúc bn học tốt
ta có: \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(A=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(A=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2017}+\frac{1}{2018}\)
\(\Rightarrow A=B\left(=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2017}+\frac{1}{2018}\right)\)
\(\Rightarrow\frac{A}{B^{2018}}=\frac{A}{A.B^{2017}}=\frac{1}{B^{2017}}\)
=> \(\frac{A}{B^{2018}}=\frac{1}{\left(\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2017}+\frac{1}{2018}\right)^{2017}}\)
ta có: \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)
\(\Rightarrow1-\frac{1}{1+2+3+...+n}=1-1:\frac{n.\left(n+1\right)}{2}=1-\frac{2}{n.\left(n+1\right)}\)
\(=\frac{n.\left(n+1\right)-2}{n.\left(n+1\right)}=\frac{n^2+n-2}{n.\left(n+1\right)}=\frac{\left(n+2\right).\left(n-1\right)}{n.\left(n+1\right)}\) (*)
Từ (*)
\(\Rightarrow1-\frac{1}{1+2}=\frac{4.1}{2.3};1-\frac{1}{1+2+3}=\frac{5.2}{3.4};...;1-\frac{1}{1+2+3+...+n}=\frac{\left(n+2\right).\left(n-1\right)}{n.\left(n+1\right)}\)
\(\Rightarrow E=\frac{4.1}{2.3}.\frac{5.2}{3.4}...\frac{\left(n+2\right).\left(n-1\right)}{n.\left(n+1\right)}=\frac{4.1.5.2...\left(n+1\right).\left(n-2\right).\left(n+2\right).\left(n-1\right)}{2.3.3.4....\left(n-1\right).n.n.\left(n+1\right)}\)\(=\frac{n+2}{n.n}\)
\(\Rightarrow\frac{E}{F}=E:F=\left(\frac{n+2}{n.n}\right):\frac{n+2}{n}=\frac{n+2}{n.n}.\frac{n}{n+2}=\frac{1}{n}\)
\(\Rightarrow\frac{E}{F}=\frac{1}{n}\)
Câu 1:
B = \(\frac{2999}{1}+\frac{2998}{2}+\frac{2997}{3}+...+\frac{1}{2999}\)
= \(\frac{3000-1}{1}+\frac{3000-2}{2}+\frac{3000-3}{3}+...+\frac{3000-2999}{2999}\)
= \(\left(\frac{3000}{1}+\frac{3000}{2}+\frac{3000}{3}+...+\frac{3000}{2999}\right)-\left(\frac{1}{1}+\frac{2}{2}+\frac{3}{3}+...+\frac{2999}{2999}\right)\)
= \(3000+3000.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2999}\right)-2999\)
= \(3000\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2999}\right)+\frac{3000}{3000}\)
= \(3000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}}{3000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}\right)}=\frac{1}{3000}\)
\(E=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)....\left(1-\frac{1}{1+2+3+...+2022}\right)\)
\(=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right).....\left(1-\frac{1}{\left(2022+1\right).2022:2}\right)\)
\(=\frac{2}{3}.\frac{5}{6}.....\frac{2023.2022:2-1}{2023.2022:2}\)
\(=\frac{4}{6}.\frac{10}{12}....\frac{\left(2023.2022:2-1\right).2}{2023.2022}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}....\frac{2021.2024}{2022.2023}\)
\(=\frac{1.2.3.4....2021}{2.3.4....2022}.\frac{4.5.6....2024}{3.4.5......2023}\)
\(=\frac{1}{2022}.\frac{2024}{3}\)
\(=\frac{1012}{3033}\)