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=>A=\(\frac{7}{2}\)(\(\frac{1}{1}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+...+\(\frac{1}{99}\)-\(\frac{1}{101}\))
=>A=\(\frac{7}{2}\)(1-\(\frac{1}{101}\))
=>A=\(\frac{350}{101}\)
7/2 ( \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{99}-\frac{1}{101}\))
7/2 ( 1 - 1/101 )
7/2 x 100/101
=350/101
\(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{99.101}\)
\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=2.\left(1-\frac{1}{101}\right)\)
\(=2.\frac{100}{101}=\frac{200}{101}\)
Đặt \(A=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+..+\frac{4}{99.101}\)
\(A=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(A=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=2.\left(1-\frac{1}{101}\right)\)
\(A=\frac{2.100}{101}=\frac{200}{101}\)
Ủng hộ mk nha !!! ^_^
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
\(=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)
\(=2.\left(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\right)\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{101}{303}-\frac{3}{303}=\frac{98}{303}\)
Đặt A = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
\(\Leftrightarrow A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.100}\)
\(=1-\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}-\frac{1}{7}+\frac{1}{9}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)
\(a)\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{-11.13^7}{11^5.13^8}=\frac{-1}{11^4.13}\) (Bạn xem thử xem có sai đề không nhé)
\(c)\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3+1\right)}{2^9.3^{10}}=\frac{2.4}{3}=\frac{8}{3}\)
\(d)\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(5.4+9\right)}=\frac{8}{20+9}=\frac{8}{29}\)
\(a)\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{\left(-11\right)\cdot13^7}{11^5\cdot13^8}=\frac{-1}{11^4\cdot13}=\frac{-1}{14641\cdot13}=\frac{-1}{190333}\)
\(c)\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^{10}\left(3^{10}-3^9\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\cdot2}{2^9\cdot3^{10}}=\frac{2\cdot2}{3}=\frac{4}{3}\)
Bài làm
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{x.\left(x+2\right)}=\frac{2015}{2016}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2015}{2016}\)
\(1-\frac{1}{x+2}=\frac{2015}{2016}\)
\(\frac{1}{x+2}=\frac{1}{2016}\)
\(\Rightarrow x+2=2016\)
\(x=2014\)
1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6+1/7-1/7+1/8-1/8+1/9+1/9-1/10
=1/2-1/10
=5/10-1/10
=4/10=2/5
\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{8x9}+\frac{1}{9x10}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\frac{1}{2}-\frac{1}{10}\)
\(\frac{2}{5}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
\(=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(=2.\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=1-\frac{1}{100}\Rightarrowđpcm\)
Ta có :
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}< 1\)\(\left(đpcm\right)\)