\(=\frac{6}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-............+\frac{1}{97}-\frac{1}{99}\right).\\ \)

\(=\frac{6}{2}\left(1-\frac{1}{97}\right)\)

tới đây tính máy là ra luôn

\(a,\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{43.45}=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{43.45}\right)=\frac{5}{3}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{45}\right)=\frac{5}{3}.\frac{44}{45}=\frac{44}{27}\)

/ là chia hay là phần vậy ạ

*Ngọc Anhh* Giống nhau cả mà bạn

Ta có:

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{199.201}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{199.201}\right)\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{199}-\frac{1}{201}\right)\)

\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{201}\right)=\frac{1}{2}.\frac{200}{201}=\frac{100}{201}\)

Còn bài kế tiếp mình không rõ quy luật nên không có giúp bạn được.

1/2 x (6/1-6/3+6/3-6/5+ ... +6/37-6/39)

1/2 x (6/1-6/39)

1/2 x 228/39

228/78

\(\frac{6}{1\cdot3}+\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{35.37}+\frac{6}{37.39}\) 

\(=3\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{37}-\frac{1}{39}\right)\) 

\(=3\left(1-\frac{1}{39}\right)=3\cdot\frac{38}{39}=\frac{114}{39}\) 

Ps: Bạn tự rút gọn nhé!!!

\(\frac{6}{1.3}+\frac{6}{3.5}+...+\frac{6}{35.37}+\frac{6}{37.39}\)

\(=3\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{35.37}+\frac{2}{37.39}\right)\)

\(=3\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{37}-\frac{1}{39}\right)\)

\(=3\left(1-\frac{1}{39}\right)\)

\(=3.\frac{38}{39}\)

\(=\frac{38}{13}\)

\(A=\frac{1}{1.3}+\frac{1}{3.5}+..+\frac{1}{97.99}\)

\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)

 \(A=\frac{1}{2}\left(1-\frac{1}{99}\right)\)

\(A=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)

\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)

\(\Leftrightarrow A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)\)

\(\Leftrightarrow A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Leftrightarrow A=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)

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