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\(\left(1-\dfrac{28}{10}\right)\left(1-\dfrac{52}{22}\right)\left(1-\dfrac{80}{36}\right)...\left(1-\dfrac{21808}{10900}\right)\)
\(=\left(1-\dfrac{2.10+8}{10}\right)\left(1-\dfrac{2.22+8}{22}\right)\left(1-\dfrac{2.36+8}{36}\right)...\left(1-\dfrac{2.10900+8}{10900}\right)\)
\(=\left(1-2-\dfrac{8}{10}\right)\left(1-2-\dfrac{8}{22}\right)\left(1-2-\dfrac{8}{36}\right)...\left(1-2-\dfrac{8}{10900}\right)\)
\(=\left(-1-\dfrac{8}{1.10}\right)\left(-1-\dfrac{8}{2.11}\right)\left(-1-\dfrac{8}{3.12}\right)...\left(-1-\dfrac{8}{100.109}\right)\)
\(=\left(\dfrac{-18}{1.10}\right)\left(\dfrac{-30}{2.11}\right)\left(\dfrac{-44}{3.12}\right)...\left(\dfrac{-10908}{100.109}\right)\)
\(=\left(\dfrac{-2.9}{1.10}\right)\left(\dfrac{-3.10}{2.11}\right)\left(\dfrac{-4.11}{3.12}\right)...\left(\dfrac{-101.108}{100.109}\right)\)
\(=\dfrac{\left(-2\right)\left(-3\right)\left(-4\right)...\left(-101\right)}{1.2.3...100}.\dfrac{9.10.11...108}{10.11.12...109}\) (1)
\(=\dfrac{101}{1}.\dfrac{9}{109}=\dfrac{909}{109}\)
Do ở (1) có \(-2-\left(-101\right)+1=100\) nhân tử (số nhân tử là số chẵn) mang dấu âm nên kết quả sẽ mang dấu dương
\(A=1+\dfrac{\dfrac{\left(1+2\right).2}{2}}{2}+\dfrac{\dfrac{\left(1+3\right).3}{2}}{3}+...+\dfrac{\dfrac{\left(1+2013\right).2013}{2}}{2013}\)
\(A=1+\dfrac{\dfrac{3.2}{2}}{2}+\dfrac{\dfrac{4.3}{2}}{3}+...+\dfrac{\dfrac{2014.2013}{2}}{2013}\)
\(A=1+\dfrac{3}{2}+\dfrac{2.3}{3}+...+\dfrac{1007.2013}{2013}\)
\(A=1+\dfrac{3}{2}+2+\dfrac{5}{2}...+1007\)
\(2A=2+3+4+5+6+...+2012+2013+2014\)
\(2A=\dfrac{\left(2+2014\right).2013}{2}\)
\(A=\dfrac{2016.2013}{4}=504.2013\)
\(B=\dfrac{-2}{1.3}+\dfrac{-2}{2.4}+...+\dfrac{-2}{2012.2014}+\dfrac{-2}{2013.2015}\)
\(-B=\dfrac{2}{1.3}+\dfrac{2}{2.4}+...+\dfrac{2}{2012.2014}+\dfrac{2}{2013.2015}\)
\(-B=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2013.2015}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2012.2014}\right)\)
\(-B=\left(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{2015-2013}{2013.2015}\right)+\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2014-2012}{2012.2014}\right)\)
\(-B=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{2013}-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2014}\right)\)
\(-B=\left(1-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2014}\right)\)
\(-B=\dfrac{2014}{2015}+\dfrac{2012}{2014.2}=\dfrac{2014^2+1006.2015}{2015.2014}\)
\(B=\dfrac{2014^2+1006.2015}{-2015.2014}\)
Đặt \(B=A\div C\)
\(C=2012+\dfrac{2011}{2}+...+\dfrac{1}{2012}=2012+\dfrac{2013-2}{2}+\dfrac{2013-3}{3}+...+\dfrac{2013-2012}{2012}\)
\(C=2012+\dfrac{2013}{2}+\dfrac{2013}{3}+...+\dfrac{2013}{2012}-1-1-...-1\)
\(C=2012+2013\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}\right)-2011\)
\(C=1+2013\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}\right)=\dfrac{2013}{2013}+2013\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}\right)\)
\(C=2013\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)=2013.A\)
\(\Rightarrow B=\dfrac{A}{C}=\dfrac{1}{2013}\)
\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)\cdot x=\left(1+\dfrac{2011}{2}\right)+\left(1+\dfrac{2010}{3}\right)+...+\left(\dfrac{1}{2012}+1\right)+1\)
\(\Leftrightarrow x\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)=\dfrac{2013}{2}+\dfrac{2013}{3}+...+\dfrac{2013}{2013}\)
=>x=2013
1/ \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}\)
\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{8}< 1\)
\(B< 1\)
2/ \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{20}\right)\)
\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{19}{20}\)
\(B=\dfrac{1\times2\times3\times...\times19}{2\times3\times4\times...\times20}\)
\(B=\dfrac{1}{20}\)
3/ \(A=\dfrac{7}{4}\cdot\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{3.4}+\dfrac{33}{4.5}+\dfrac{33}{5.6}+\dfrac{33}{6.7}\right)\)
\(A=\dfrac{7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}.\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}\cdot\dfrac{4}{21}\)
\(A=11\)
4/ A phải là \(\dfrac{2011+2012}{2012+2013}\)
Ta có : \(B=\dfrac{2011}{2012}+\dfrac{2012}{2013}>\dfrac{2011}{2013}+\dfrac{2012}{2013}=\dfrac{2011+2012}{2013}>\dfrac{2011+2012}{2012+2013}=A\)
\(\Rightarrow B>A\)
\(a)\left(5^{2010}+5^{2012}+5^{2014}\right):\left(5^{2011}+5^{2009}+5^{2007}\right)\)
\(=\dfrac{5^{2007}\left(5^3+5^5+5^7\right)}{5^{2007}\left(5^4+5^2+1\right)}=\dfrac{5^3+5^5+5^7}{5^4+5^2+1}\)
\(=\dfrac{125+3125+78125}{625+25+1}=\dfrac{81375}{651}=125\)
\(b)-\dfrac{7}{45}+\dfrac{1}{4}+\dfrac{3}{5}+\dfrac{1}{12}+\dfrac{2}{3}+\dfrac{1}{39}+\dfrac{5}{9}\)
\(=\dfrac{-7.52+1.585+3.468+1.195+2.780+1.60-5.260}{2340}\)
\(=\dfrac{-364+585+1404+195+1560+60-1300}{2340}\)
\(=\dfrac{2140}{2340}=\dfrac{107}{117}\)
Câu Q:Tương tự như mấy câu khác thôi