Tính:
.\(\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}\)
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27 tháng 9 2016
Đặt \(A=\frac{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}\)
Tử số của A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
\(\Rightarrow A=1\left(đpcm\right)\)
29 tháng 6 2021
\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(B=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(B< \frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)
\(B< \frac{50}{60}\Leftrightarrow B< \frac{5}{6}\)
Ta có: \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)(1)
Thay (1) vào đề bài
\(\Rightarrow\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}=1\)
\(\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}\)
\(=1\)