\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\)

\(A=2.\left(\frac{1}{2.6}+\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{198.202}\right)\)

\(A=2.\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{198}-\frac{1}{202}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{202}\right)\)

\(A=\frac{1}{2}.\frac{50}{201}\)

\(A=\frac{25}{101}\)

=25/101

k cho to nhe

\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\)

\(A=2.\left(\frac{1}{2.6}+\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{198.202}\right)\)

\(A=2.\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{198}-\frac{1}{202}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{202}\right)\)

\(A=\frac{1}{2}.\frac{50}{201}\)

\(A=\frac{25}{101}\)

đề sai rồi

đề sai rồi

\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\)

\(A=2.\left(\frac{1}{2.6}+\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{198.202}\right)\)

\(A=2.\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{198}-\frac{1}{202}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{202}\right)\)

\(A=\frac{1}{2}.\frac{50}{201}\)

\(A=\frac{25}{101}\)

\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\)

\(A=2.\left(\frac{1}{2.6}+\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{198.202}\right)\)

\(A=2.\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{198}-\frac{1}{202}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{202}\right)\)

\(A=\frac{1}{2}.\frac{50}{201}\)

\(A=\frac{25}{101}\)

 Như bạn kia là rất đúng 

\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\)

\(A=\frac{1}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

\(4A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(4A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\)

\(4A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)

\(4A=1-\frac{1}{101}=\frac{100}{101}\)

\(A=\frac{100}{101.4}=\frac{25}{101}\)

\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+\frac{1}{14.9}+...+\frac{1}{198.101}\)

\(A=\frac{1}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)

\(4A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)

\(4A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)

\(4A=1-\frac{1}{101}=\frac{100}{101}\)

\(A=\frac{100}{101}:4=\frac{25}{101}\)

đáp số 

25/101

hok tốt

200000+200000=?

200000+200000=400000

chi tiet

tất cả rút \(\frac{1}{2}\) ra ngoài ta có :

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

đến đây thì dễ rồi tự làm tiếp đi , ko hiểu thì hỏi nha

cái này bn đặt làm hiệu sẽ ra ngay thôi!