\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)

\(2A=\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{51-49}{49.51}\)

\(2A=\frac{5}{3.5}-\frac{3}{3.5}+\frac{7}{5.7}-\frac{5}{5.7}+...+\frac{51}{49.51}-\frac{49}{49.51}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

\(2A=\frac{1}{3}-\frac{1}{51}=\frac{27}{51}-\frac{1}{51}=\frac{26}{51}\)

\(A=\frac{26}{51}:2=\frac{13}{51}\)

Đặt   \(A=\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{49.51}\)

\(\Rightarrow2A=2\left(\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{49.51}\right)\)

\(\Rightarrow2A=\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{49.51}\)

\(\Rightarrow2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{49}-\frac{1}{51}\)

\(\Rightarrow2A=\frac{1}{3}-\frac{1}{51}\)

\(\Rightarrow2A=\frac{16}{51}\)

\(\Rightarrow A=\frac{16}{51}:2\)

\(\Rightarrow A=\frac{8}{51}\)

Vậy \(\Rightarrow\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{49.51}=\frac{8}{51}\)

goi A là tổng

2A=2/3.5+2/5.7+...+2/49.51

2A=2/3-2/5+2/5.2/7+...+2/49-2/51

2A=2/3-2/51            2A=32/51             A=32/51:2             A=16/51

=\(\frac {1} {2}(\frac {2}{3.5}+\frac {2}{5.7}+...+\frac {2}{49.51})\)=\(\frac {1}{2}(\frac {1}{3}-\frac {1}{5}+\frac {1}{5}-\frac {1}{7}+...+\frac {1}{49}-\frac {1}{51})\)=\(\frac {1}{2}(\frac {1}{3} -\frac {1}{51})\)=\(\frac {1}{2}.\frac {16}{51}\)=\(\frac {8}{51}\)

\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{49\cdot51}\)

\(=2\left(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{49\cdot51}\right):2\)

\(=\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{49\cdot51}\right):2\)

\(=\left(\frac{1}{3}-\frac{1}{51}\right):2\)

\(=\frac{16}{51}:2\)

\(=\frac{8}{51}\)

\(=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right):2\)

bài mk dòng cuối cùng bỏ đi bạn nhé