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

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

\(=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)

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

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)

\(\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+....+\frac{3}{49.51}\)

= 3. \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{49}-\frac{1}{51}\right)\)

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

=\(\frac{3}{2}.\frac{16}{51}\)

=\(\frac{8}{17}\)

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\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

~ Hok tốt ~

\(\)

Viết thành 2 . (1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/97.99

Tui hk bít nữa

Bài này bạn nhân 2B ra

Sau đó tách mỗi phân số thành 2 hiệu

Từ đó triệt tiêu sẽ ra 2 số cuối cùng

Bạn trừ 2 số đó với nhau là ra

P/s : Sorry mình đg ôn thi nên ko nên giải trực tiếp, thông cảm nha

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

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{16}{2.51}=\frac{8}{51}\)

.

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

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)

\(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{49.51}\)

\(=\frac{1}{2}.\left(\frac{2}{5}.7+\frac{2}{7}.9+\frac{2}{9}.11+...+\frac{2}{49}.51\right)\)

\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{51}\right)\)

\(=\frac{1}{2}.\frac{46}{255}\)

\(=\frac{23}{255}\)

\(\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{49.51}\)

\(\Rightarrow2 \left(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{49.51}\right)\)

\(\Rightarrow\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{49}-\frac{1}{51}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{51}=\frac{46}{255}\)

Vì biểu thức đã được nhân 2 nên giá trị của biểu thức là:

\(\frac{46}{255}:2=\frac{23}{255}\)

\(\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{49.51}\)

\(=\frac{1}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{49.51}\right)\)

\(=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{51}\right)\)

\(=\frac{1}{2}\cdot\frac{46}{255}\)

\(=\frac{23}{255}\)

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

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

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

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

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

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

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

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

NHẦM GIẢI LẠI :

\(A=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{3}{2}.\frac{16}{51}=\frac{8}{17}\)