\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\frac{98}{99}\)
\(=\frac{49}{99}\)
1/1.3+1/3.5+...+1/97.99
=(2/1.3+2/3.5+...+2/97.99):2
=(1-1/3+1/3-1/5+...+1/97-1/99):2
=(1-1/99):2
=99-1/99.2
=49/99
nhớ cho mk nha
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)
Suy ra: \(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
\(2A=1-\frac{1}{99}\)
\(2A=\frac{98}{99}\)
\(A=\frac{98}{99}.\frac{1}{2}=\frac{49}{99}\)
