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\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\)
\(A=\frac{1}{3}-\frac{1}{101}\)
\(A=\frac{98}{303}\)
A = \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{99.101}\)
= \(\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{99.101}\right):2\)\(=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right):2\)
= \(\left(\frac{1}{3}-\frac{1}{101}\right):2=\frac{101-3}{303}:2=\frac{98}{303}:2=\frac{49}{303}\)
Dấu chấm trong bài là dấu nhân nha !
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1/3x5 +1/5x7+1/7x9 +1/9x11+...+1/99x101
1/3-1/5+1/5-1/7+...+1/99-1/101
1/3-1/101
98/303
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\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\)
\(\Rightarrow\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(\Rightarrow\frac{1}{2}.\frac{98}{303}\)
\(\Rightarrow\frac{49}{303}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
49/303,xin lỗi bạn mk làm biếng viết lời giải nếu cần nói mk nha
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a=lộn