\(A=\frac{1}{2.1.3}+\frac{1}{2.3.5}+\frac{1}{2.5.7}+...+\frac{1}{2.99.101}\)

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

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)

\(=\frac{1}{2}.\frac{100}{101}\)

\(=\frac{50}{101}\)

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

Đặt :\(M=\frac{1}{2.6}+\frac{1}{6.10}+...+\frac{1}{194.198}\)

\(M=\frac{1}{4}\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+...+\frac{1}{194}-\frac{1}{198}\right)\)

\(M=\frac{1}{4}\left(\frac{1}{2}-\frac{1}{198}\right)\)

\(M=\frac{1}{4}.\frac{49}{99}\)

\(M=\frac{49}{396}\)

Đặt \(N=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(N=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(N=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(N=\frac{1}{2}.\frac{98}{303}\)

\(N=\frac{49}{303}\)

Vậy ta có : A = M + N = \(\frac{49}{396}+\frac{49}{303}\) , bạn tự tính luôn nha

đề sai rồi

đề sai rồi

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

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

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

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

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

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

=25/101

k cho to nhe

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

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

Ta thấy : thừa số thứ nhất ở mẫu của phân số liền sau = thừa số thứ nhất của phân số liền trước + 4

Thừa số thứ hai ở mẫu của phân số liền sau = thừa số thứ hai của phân số liền trước + 2 

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

\(4A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{101-99}{99.101}\)

4A= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)

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

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

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

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

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

\(A=\frac{1}{2}\times\frac{50}{101}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

 Như bạn kia là rất đúng 

=>A:1/2=1/1x3+1/3x5+1/5x7+...+1/99x101

=>2a=1/2(2/1x3+2/3x5+...+2/99x101)

từ đây tự làm

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

\(\Rightarrow2A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(\Rightarrow2A=\frac{1}{2}\left(1-\frac{1}{101}\right)\)

\(\Rightarrow4A=\frac{100}{101}\)

\(\Leftrightarrow A=\frac{100}{101}.\frac{1}{4}=\frac{4.25}{101.4}=25< 26\)