\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2006}}\)
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}\)
\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2006}}\right)\)
\(A=2-\frac{1}{2^{2006}}\)
\(ChoM=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{999}\\ N=\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}\\ Tính\frac{M}{N}\)
Tính giá trị biểu thức sau: \(C=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}}\)
Tính nhanh \(C=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}}\)
Giải chi tiết cho mk nha, thanks!
Tính A=\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{999.1}}\)
\(\frac{1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+....\frac{1}{997.3}+\frac{1}{991.1}}\)
\(C=\frac{1}{1.999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999.1}\)
\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+......+\frac{1}{999}}{\frac{1}{1.999}+\frac{1}{3.997}+\frac{1}{5.995}+......+\frac{1}{999.1}}\)
\(B=\frac{1+\left(1+2\right)+\left(1+2+3\right)+\left(1+2+3+4\right)+......+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+4.5+......+98.99}\)
\(C=\frac{\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+......+\frac{1}{100.400}}{\frac{1}{1.102}+\frac{1}{2.103}+\frac{1}{3.104}+......+\frac{1}{299.400}}\)
\(D=\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+......+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{100}}:\frac{92-\frac{1}{9}-\frac{2}{10}-\frac{3}{97}-......-\frac{92}{100}}{\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+......+\frac{1}{500}}\)
Tìm x biết: \(\frac{3}{2x+1}\)+ \(\frac{10}{4x+2}\)- \(\frac{6}{6x+3}\)=\(\frac{12}{26}\)
Chứng minh rằng: \(\frac{1}{5^2}\)+ \(\frac{1}{6^2}\)+...+ \(\frac{1}{2007^2}\)> \(\frac{1}{5}\)
rút gọn: M = ( 1+\(\frac{1}{3}\)+ \(\frac{1}{5}\)+ ... + \(\frac{1}{999}\)) : (\(\frac{1}{1.999}\)+ \(\frac{1}{3.997}\)+...+\(\frac{1}{997.3}\)+\(\frac{1}{999.1}\))
Giúp mình nha!!!
Bài 1: Cho các số tự nhiên n1;n2;...;n10 thỏa mãn \(N=n_1+n_2+...+n_{10}=2013\)
Đặt \(S=n_1^2+n_2^2+...+n_{10}^2\)
Chứng minh S-1 chia hết cho 2
Bài 2: Tính
\(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right):\left(\frac{1}{999}+\frac{1}{3.997}+...+\frac{1}{997.3}+\frac{1}{999}\right)\)
