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\(\dfrac{1}{3}\)x(\(\dfrac{3}{1+4}\)+\(\dfrac{3}{4+7}\)+........+\(\dfrac{3}{101+103}\))
\(\dfrac{1}{3}\)x(\(\dfrac{1}{1}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+.........+\(\dfrac{ }{ }\)\(\dfrac{1}{101}\)-\(\dfrac{1}{103}\))
\(\dfrac{1}{3}\)x(\(\dfrac{1}{1}\)-\(\dfrac{1}{103}\))
\(\dfrac{1}{3}\)x\(\dfrac{102}{103}\)
\(\dfrac{34}{103}\)
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)
\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(=\dfrac{1}{1}-\dfrac{1}{103}\)
\(=\dfrac{102}{103}\)
\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{97}-\dfrac{1}{100}\right)=\dfrac{0,33x}{2009}\)
\(\Leftrightarrow\dfrac{1}{3}\cdot\dfrac{99}{100}=\dfrac{0,33x}{2009}\)
\(\Leftrightarrow\dfrac{33}{100}=\dfrac{0,33x}{2009}\) <=> x = (tự tính )
⇔13(11−14+14−...+197−1100)=0,33x2009⇔13(11−14+14−...+197−1100)=0,33x2009
⇔13⋅99100=0,33x2009⇔13⋅99100=0,33x2009
\(a,A=\dfrac{101}{100}+\dfrac{102}{100}+\dfrac{103}{100}+...+\dfrac{199}{100}\)
\(A=\dfrac{101+102+103+...+109}{100}\)
Xét tử số : \(101+102+103+...+199\)
Có : \(\left(199-101\right):1+1=99\) (số hạng)
\(\Rightarrow\) Tử số bằng \(:\left(199+101\right).99:2=14850\)
\(\Rightarrow A=\dfrac{14850}{100}=\dfrac{297}{2}\)
\(b,B=\dfrac{10002}{10000}+\dfrac{10004}{10000}+\dfrac{10006}{10000}+...+\dfrac{12014}{10000}\)
\(B=\dfrac{10002+10004+10006+...+12014}{10000}\)
\(B=\dfrac{10002+10004+10006+...+12014}{10000}\)
Xét tử số : \(10002+10004+10006+...+12014\)
Có : \(\left(12014-10002\right):2+1=1007\) (số hạng)
\(\Rightarrow\) Tử số bằng : \(\left(12014+10002\right).1007:2=11085056\)
\(\Rightarrow B=\dfrac{11085056}{10000}\)
Bạn tự làm câu C nha
\(D=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{2014.2015}\)
\(\Rightarrow D=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\)
\(\Rightarrow D=\dfrac{1}{5}-\dfrac{1}{2015}=\dfrac{402}{2015}\)
\(E=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{2014.2017}\)
\(\Rightarrow3E=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{2014.2017}\)
\(\Rightarrow3E=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2014}-\dfrac{1}{2017}\)
\(\Rightarrow3E=1-\dfrac{1}{2017}=\dfrac{2016}{2017}\)
\(\Rightarrow E=\dfrac{2016}{2017}:3=\dfrac{672}{2017}\)
D = \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\) +...+ \(\dfrac{1}{2014.2015}\)
D = \(\dfrac{1}{5}\) - \(\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)+...+ \(\dfrac{1}{2014}-\dfrac{1}{2015}\)
D = \(\left(\dfrac{1}{5}-\dfrac{1}{2015}\right)\)
D = \(\dfrac{403}{2015}-\dfrac{1}{2015}\)
D = \(\dfrac{402}{2015}\)
A = \(-\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2017}-\dfrac{1}{2020}\right)=-1+\dfrac{1}{2020}=\dfrac{-2019}{2020}\)
b: Ta có: \(B=\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{64\cdot69}\)
\(=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{64\cdot69}\right)\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{64}-\dfrac{1}{69}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{65}{4\cdot69}\)
\(=\dfrac{13}{276}\)
\(A=\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+...+\dfrac{2}{97\cdot100}\\ A=\dfrac{2}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{97\cdot100}\right)\\ A=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\\ A=\dfrac{2}{3}\left(1-\dfrac{1}{100}\right)=\dfrac{2}{3}\cdot\dfrac{99}{100}=\dfrac{33}{50}\\ B=\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{64\cdot69}\\ B=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{64\cdot69}\right)\\ B=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{64}-\dfrac{1}{69}\right)\\ B=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{69}\right)=\dfrac{1}{5}\cdot\dfrac{65}{276}=\dfrac{13}{276}\)
\(C=70\left(\dfrac{13}{56}+\dfrac{13}{72}+\dfrac{13}{90}\right)=70\cdot13\left(\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\\ C=910\left(\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\\ C=910\left(\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ C=910\left(\dfrac{1}{7}-\dfrac{1}{10}\right)=910\cdot\dfrac{3}{70}=39\)
a) \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{2018}{2019!}\\ =\left(\dfrac{1}{1!}-\dfrac{1}{2!}\right)+\left(\dfrac{1}{2!}-\dfrac{1}{3!}\right)+...+\left(\dfrac{1}{2018!}-\dfrac{1}{2019!}\right)\\ =1-\dfrac{1}{2019!}< 1\)
b) \(\dfrac{1\cdot2-1}{2!}+\dfrac{2\cdot3-1}{3!}+...+\dfrac{999\cdot1000-1}{1000!}\\ =\dfrac{1\cdot2}{2!}-\dfrac{1}{2!}+\dfrac{2\cdot3}{3!}-\dfrac{1}{3!}+...+\dfrac{999-1000}{1000!}-\dfrac{1}{1000!}\\ =\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{1!}-\dfrac{1}{3!}+\dfrac{1}{2!}-\dfrac{1}{4!}+...+\dfrac{1}{999!}+\dfrac{1}{1000!}\\ =1+1-\dfrac{1}{1000!}\\ =2-\dfrac{1}{1000!}< 2\)
a: \(A=6\left(x+\dfrac{1}{3}\right)^2-7>=-7>-8\forall x\)
\(B=-8-\left(3.75-x\right)^2\le-8\)
Do đó: A>B
b: \(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}=\dfrac{15}{16}\)
\(B=\left(\dfrac{1}{2}\right)^4=\dfrac{1}{16}\)
Do đó: A>B
\(l=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+....+\dfrac{1}{97.100}\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{100}\right)=\dfrac{1}{3}-\dfrac{1}{300}< \dfrac{1}{3}\left(đpcm\right)\)