Đặt S là biểu thức trên
\(\Rightarrow S=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+........+\frac{2}{97.99}\right)\)
\(\Rightarrow S=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-.........-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow S=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(\Rightarrow S=\frac{1}{2}\left(\frac{99}{99}-\frac{1}{99}\right)\)
\(\Rightarrow S=\frac{1}{2}.\frac{98}{99}\)
\(\Rightarrow S=\frac{49}{99}\)
Vậy biểu thức trên có giá trị là \(\frac{49}{99}\)
\(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{97\times99}\)
\(=\frac{1}{2}\times\left(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+....+\frac{1}{97\times99}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}\times\frac{98}{99}\)
\(=\frac{49}{99}\)
\(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{97x99}\)
\(=\frac{1}{2}x(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99})\)
\(=\frac{1}{2}x\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}x\frac{98}{99}\)
\(=\frac{49}{99}\)
\(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{97\times99}\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}\times\frac{98}{99}\)
\(=\frac{49}{99}\).
~ HOK TỐT ~
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\frac{98}{99}\)
\(=\frac{49}{99}\)