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#)Giải :
Ta có : \(\frac{101}{2}.\frac{102}{2}.\frac{103}{2}.....\frac{200}{2}=\frac{101.102.103.....200}{2^{100}}=\frac{\left(101.102.103.....200\right)\left(1.2.3.....100\right)}{2^{100}\left(1.2.3.....100\right)}\)
\(=\frac{1.2.3.....200}{\left(2.1\right)\left(2.2\right)\left(2.3\right)...\left(2.100\right)}=\frac{\left(1.3.5.....99\right)\left(2.4.6.....100\right)}{2.4.6.....200}=1.3.5.....99\left(đpcm\right)\)
Ta có : 1.3.5.7.....199 = \(\frac{\left(1.3.5.7.....199\right).\left(2.4.6.8.....200\right)}{2.4.6.8.....200}=\frac{1.2.3.4.5.....199.200}{\left(1.2\right).\left(2.2\right).\left(3.2\right).....\left(100.2\right)}=\frac{1.2.3.4.5.....199.200}{2^{100}.1.2.3.....100}=\frac{101.102.103.....200}{2^{100}}\)\(=\frac{101}{2}.\frac{102}{2}\frac{103}{2}.....\frac{200}{2}\)\( \left(ĐPCM\right)\)
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1.3.5.....197.199 = \(\frac{\left(1.3.5.....197.199\right)\left(2.4.6.....198.200\right)}{2.4.6......198.200}\)= \(\frac{1.2.3......199.200}{2^{100}.\left(1.2.3.....100\right)}=\frac{101.102.103......200}{2^{100}}=\frac{101}{2}.\frac{102}{2}.\frac{103}{2}.....\frac{200}{2}\)
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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
\(\Rightarrow\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}=1\)
\(\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}\)
\(=1\)