\(\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{2}{7}\right)\left(1-\dfrac{3}{7}\right)...\left(1-\dfrac{2014}{7}\right)\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
2: \(=\dfrac{203}{60}\cdot\dfrac{81}{1225}=\dfrac{783}{3500}\)
\(G=\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{2}{7}\right)\left(1-\dfrac{3}{7}\right)\cdot...\cdot\left(1-1\dfrac{2}{7}\right)\left(1-1\dfrac{3}{7}\right)\\ =\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{2}{7}\right)\left(1-\dfrac{3}{7}\right)\cdot...\left(1-\dfrac{7}{7}\right)...\left(1-1\dfrac{2}{7}\right)\left(1-1\dfrac{3}{7}\right)\\ =\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{2}{7}\right)\left(1-\dfrac{3}{7}\right)\cdot...\cdot0\cdot...\cdot\left(1-1\dfrac{2}{7}\right)\left(1-1\dfrac{3}{7}\right)\\ =0\\ VậyG=0\)
\(G=\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{2}{7}\right)\left(1-\dfrac{3}{7}\right)...\left(1-1\dfrac{2}{7}\right)\left(1-1\dfrac{3}{7}\right)\)
\(\Leftrightarrow G=\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{2}{7}\right)\left(1-\dfrac{3}{7}\right)....\left(1-\dfrac{7}{7}\right)...\left(1-1\dfrac{2}{7}\right)\left(1-1\dfrac{3}{7}\right)\)
\(\Leftrightarrow G=\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{2}{7}\right)\left(1-\dfrac{3}{7}\right)...0....\left(1-1\dfrac{2}{7}\right)\left(1-1\dfrac{3}{7}\right)\)
\(\Leftrightarrow G=0\)
Vậy \(G=0\)
\(S=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+....+\left(-\dfrac{1}{7}\right)^{2017}\\ =1+-\dfrac{1}{7}+\dfrac{1}{7^2}+-\dfrac{1}{7^3}+.....+-\dfrac{1}{7^{2017}}\\ =\left(1+\dfrac{1}{7^2}+\dfrac{1}{7^4}+...+\dfrac{1}{7^{2016}}\right)-\left(\dfrac{1}{7}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{2017}}\right)\)
rồi bạn tính 2 về rồi trừ ra là xng nhé
Các bạn trả lời giúp mk nha. Mk đang cần gấp. Chều nay mk kiểm tra rồi
a) \(\left(\dfrac{2}{3}-\dfrac{1}{2}-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}-\dfrac{1}{7}\right)\)
\(=-\dfrac{1}{6}\cdot\dfrac{17}{28}\)
\(=-\dfrac{17}{168}\)
b) \(\left(\dfrac{15}{21}\div\dfrac{5}{7}\right)\div\left(\dfrac{6}{5}\div2\right)\)
\(=1\div\dfrac{3}{5}\)
\(=\dfrac{5}{3}\)
\(B=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2018}\)
\(\Rightarrow-\dfrac{1}{7}B=\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+\left(-\dfrac{1}{7}\right)^3+...+\left(-\dfrac{1}{7}\right)^{2019}\)
\(\Rightarrow-\dfrac{1}{7}B-1=\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+\left(-\dfrac{1}{7}\right)^3+...+\left(-\dfrac{1}{7}\right)^{2019}-\left(-\dfrac{1}{7}\right)^0-\left(-\dfrac{1}{7}\right)^1-\left(-\dfrac{1}{7}\right)^2-...-\left(-\dfrac{1}{7}\right)^{2018}\)
\(\Rightarrow-\dfrac{8}{7}B=\left(-\dfrac{1}{7}\right)^{2019}-1\)
\(\Rightarrow B=\left[\left(-\dfrac{1}{7}\right)^{2019}-1\right]:\left(-\dfrac{8}{7}\right)\)
\(B=1-\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}+...-\dfrac{1}{7^{2017}}+\dfrac{1}{7^{2018}}\\ \Rightarrow7B=7-1+\dfrac{1}{7}-\dfrac{1}{7^2}+...-\dfrac{1}{7^{2016}}+\dfrac{1}{7^{2017}}\\ \Rightarrow7B+B=6+\dfrac{1}{7}-\dfrac{1}{7^2}+...+\dfrac{1}{7^{2017}}+1-\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}+...-\dfrac{1}{7^{2017}}+\dfrac{1}{7^{2018}}\\ \Rightarrow8B=7+\dfrac{1}{7^{2018}}=\dfrac{7^{2019}+1}{7^{2018}}\\ \Rightarrow B=\dfrac{7^{2019}+1}{8\cdot7^{2018}}\)
\(=\dfrac{6}{7}\cdot\dfrac{5}{7}\cdot\dfrac{4}{7}\cdot\dfrac{3}{7}\cdot\dfrac{2}{7}\cdot\dfrac{1}{7}\cdot\dfrac{0}{7}\cdot...\cdot\dfrac{-2007}{7}\)
=0