\(C=1-\frac{2}{2.3}+1-\frac{2}{3.4}+...+1-\frac{2}{2019.2020}\)
\(=2018-2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\right)\)
\(=2018-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)
\(=2018-2\left(\frac{1}{2}-\frac{1}{2020}\right)\)
\(=2018-2.\frac{1009}{2020}\)
\(=2018-\frac{1009}{1010}\)
\(=\frac{2037171}{1010}\)
Cho \(S_1-S_2+S_3-S_4+S_5=\frac{m}{n}\) với m, n nguyên tố cùng nhau. Biết:
\(S_1=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)
\(S_2=\frac{1}{2\cdot3}+\frac{1}{2\cdot4}+\frac{1}{2\cdot5}+\frac{1}{2\cdot6}+\frac{1}{3\cdot4}+\frac{1}{3\cdot5}+\frac{1}{3\cdot6}+\frac{1}{4\cdot5}+\frac{1}{4\cdot6}+\frac{1}{5\cdot6}\)
\(S_3=\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot5}+\frac{1}{2\cdot3\cdot6}+\frac{1}{2\cdot4\cdot5}+\frac{1}{2\cdot4\cdot6}+\frac{1}{2\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5}+\frac{1}{3\cdot4\cdot6}+\frac{1}{3\cdot5\cdot6}+\frac{1}{4\cdot5\cdot6}\)
\(S_4=\frac{1}{2\cdot3\cdot4\cdot5}+\frac{1}{2\cdot3\cdot4\cdot6}+\frac{1}{2\cdot3\cdot5\cdot6}+\frac{1}{2\cdot4\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5\cdot6}\)
\(S_5=\frac{1}{2\cdot3\cdot4\cdot5\cdot6}\)
Tính \(m+n\)
1)Giải phương trình
\(\sqrt[3]{2-x}=1-\sqrt{x-1}.\)
2)Cho \(S=\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)...\left(1-\frac{2}{2020.2021}\right)\)là một tích của 2019 thừa số. Tính S.
Tính tổng S= \(\frac{1}{1+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{2019^2}+\sqrt{2019^2-2}}\)
Tìm n, biết:
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{n\cdot\left(n+1\right)\cdot\left(n+2\right)}>0,24995\)
M = \(\sqrt{1+2019^2+\frac{2019^2}{2020^2}}+\frac{2019}{2020}\)
N = \(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2019^2}+\frac{1}{2020^2}}\)
Cám ơn các cậu.
tính B=\(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+....+\sqrt{1+\frac{1}{2018^2}+\frac{1}{2019^2}}\)
Chứng minh :
A = \(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2018^2}+\frac{1}{2019^2}}+\sqrt{1+\frac{1}{2019^2}+\frac{1}{2020^2}}\)
là 1 số hữu tỉ .
Tính tống
\(S=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}\)\(+....+\sqrt{1+\frac{1}{2018^2}+\frac{1}{2019^2}}\)
HELP!!!
Tính:
A= \(\frac{1}{2\sqrt{1}+1\sqrt{2}}\)+ \(\frac{1}{3\sqrt{2}+2\sqrt{3}}\)+....+ \(\frac{1}{2019\sqrt{2018}+2018\sqrt{2019}}\)
