Cho \(A=1 +\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2^{2023}-1}\) CMR: \(A>\dfrac{2023}{2}\)

𝓣𝓪 𝓬𝓸́: \(1-\dfrac{2002}{2003}=\dfrac{1}{2003}\)

            \(1-\dfrac{2003}{2004}=\dfrac{1}{2004}\)  

𝓓𝓸 \(\dfrac{1}{2003}>\dfrac{1}{2004}\) 

𝓷𝓮̂𝓷 \(\dfrac{2002}{2003}>\dfrac{2003}{2004}\)

𝓥𝓪̣̂𝔂 \(\dfrac{2002}{2003}>\dfrac{2003}{2004}\)

C.Ba cụm danh từ.

☘️Chúc bạn học tốt☘️

\(\left(x-20\right).\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{200}}{\dfrac{1}{199}+\dfrac{2}{198}+\dfrac{3}{197}+...+\dfrac{198}{2}+\dfrac{199}{1}}=\dfrac{1}{200}.\\ \left(x-20\right)\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{200}}{\left(1+\dfrac{1}{199}\right)+\left(1+\dfrac{2}{198}\right)+\left(1+\dfrac{3}{197}\right)+...+\left(1+\dfrac{198}{2}\right)+1}=\dfrac{1}{200}.\)

\(\left(x-20\right).\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{200}}{\dfrac{200}{199}+\dfrac{200}{198}+\dfrac{200}{197}+...+\dfrac{200}{2}+\dfrac{200}{200}}=\dfrac{1}{200}.\\ \left(x-20\right).\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{200}}{200\left(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+...+\dfrac{1}{2}+\dfrac{1}{200}\right)}=\dfrac{1}{200}.\\ \left(x-20\right).\dfrac{1}{200}=\dfrac{1}{200}.\)

\(\Rightarrow x-20=\dfrac{1}{200}:\dfrac{1}{200}=1.\\ \Rightarrow x=1+20=21.\)

\(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{9999}{10000}.\\ =\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}...\dfrac{99.101}{100^2}.\\ =\dfrac{1.3.2.4.3.5...99.101}{2^2.3^2.4^2...100^2}.\\ =\dfrac{\left(1.2.3...99\right).\left(3.4.5...101\right)}{\left(2.3.4...100\right).\left(2.3.4...100\right)}.\\ =\dfrac{\left(1.1.1...1\right).\left(1.1.1...101\right)}{\left(1.1.1...100\right).\left(2.1.1...1\right)}=\dfrac{1.101}{100.2}=\dfrac{101}{200}.\)

\(C=\left(1+\dfrac{2}{3}\right).\left(1+\dfrac{2}{5}\right).\left(1+\dfrac{2}{7}\right)...\left(1+\dfrac{2}{2009}\right).\left(1+\dfrac{2}{2011}\right).\\ C=\dfrac{5}{3}.\dfrac{7}{5}.\dfrac{9}{7}...\dfrac{2011}{2009}.\dfrac{2013}{2011}.\\ C=\dfrac{5.7.9...2011.2013}{3.5.7...2009.2011}.\\ C=\dfrac{1.1.1...1.2013}{3.1.1...1.1}.\\ C=C=\dfrac{2013}{3}=671.\)

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}.\\= \dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}.\\ =\dfrac{1}{1}+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5}\right)+...+\left(-\dfrac{1}{99}+\dfrac{1}{99}\right)-\dfrac{1}{101}.\\ =\dfrac{1}{1}+0+0+...+0-\dfrac{1}{101}.\\ =1-\dfrac{1}{101}=\dfrac{100}{101}.\)

Tìm các cặp số nguyên x và y biết: \(\dfrac{x}{4}-\dfrac{3}{y}=\dfrac{5}{8}\)