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A= \(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{9}{11}=\dfrac{1}{3}-\dfrac{7}{9}=\dfrac{3}{9}-\dfrac{7}{9}=-\dfrac{4}{9}\)
\(B=\left(\dfrac{1}{5}+\dfrac{2}{15}+\dfrac{2}{3}\right)+\left(-\dfrac{2}{7}+\dfrac{1}{42}-\dfrac{13}{28}-\dfrac{1}{4}\right)\)
\(=\dfrac{3+2+10}{15}+\dfrac{-2\cdot12+2-13\cdot3-21}{84}\)
=1-82/84
=2/84=1/42
\(C=\dfrac{1}{50}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\right)\)
\(=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(=\dfrac{1}{50}-1+\dfrac{1}{50}=\dfrac{1}{25}-1=-\dfrac{24}{25}\)
\(D=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)
Ta đặt
\(A=\dfrac{1}{50}-\dfrac{1}{50\times49}-....-\dfrac{1}{2\times1}\)
\(A=\dfrac{1}{50}-\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{49\times50}\right)\)
\(A=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(A=\dfrac{1}{50}-\left(1-\dfrac{1}{50}\right)\)
\(A=\dfrac{1}{50}-\dfrac{49}{50}\)
\(A=\dfrac{-48}{50}=\dfrac{-24}{25}\)
\(=\dfrac{1}{50}-\left(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{50-49}{49.50}\right)=\)
\(=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=\)
\(=\dfrac{1}{50}-\left(1-\dfrac{1}{50}\right)=\dfrac{2}{50}-1=\dfrac{1}{25}-1=-\dfrac{24}{25}\)
\(\dfrac{1}{2014}-\dfrac{1}{2014.2013}-\dfrac{1}{2013.2012}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}=\dfrac{1}{2014}-\left(\dfrac{1}{2013.2014}+\dfrac{1}{2012.2013}+....+\dfrac{1}{1.2}\right)=\dfrac{1}{2014}-\left(\dfrac{1}{2013}-\dfrac{1}{2014}+\dfrac{1}{2012}-\dfrac{1}{2013}+...+1-\dfrac{1}{2}\right)=\dfrac{1}{2014}-\left(1-\dfrac{1}{2014}\right)=\dfrac{1}{2014}-\dfrac{2013}{2014}=-\dfrac{2012}{2014}=-\dfrac{1006}{1007}\)
\(A=\dfrac{1}{100}-\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-...-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\\ =\dfrac{1}{100}+1=\dfrac{101}{100}\)
\(A=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(A=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(A=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)
\(A=\dfrac{1}{100}-\dfrac{99}{100}=\dfrac{-49}{50}\)
\(F=-\dfrac{1}{1.2}-\dfrac{1}{2.3}-...-\dfrac{1}{2014.2015}-\dfrac{1}{2015.2016}\)
\(\Rightarrow-F=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}+\dfrac{1}{2015}-\dfrac{1}{2016}=1-\dfrac{1}{2016}=\dfrac{2015}{2016}\)\(\Rightarrow F=\dfrac{-2015}{2016}\)
Giải:
\(F=\dfrac{-1}{2016.2015}-\dfrac{1}{2015.2014}-\dfrac{1}{2014.2013}-\dfrac{1}{2013.2012}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(\Leftrightarrow F=-\left(\dfrac{1}{2016.2015}+\dfrac{1}{2015.2014}+\dfrac{1}{2014.2013}+\dfrac{1}{2013.2012}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)
\(\Leftrightarrow F=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2012.2013}+\dfrac{1}{2013.2014}+\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}\right)\)
\(\Leftrightarrow F=-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)
\(\Leftrightarrow F=-\left(\dfrac{1}{1}-\dfrac{1}{2016}\right)\)
\(\Leftrightarrow F=-\dfrac{2015}{2016}\)
Vậy ...
Giải:
\(\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(=\dfrac{1}{99}-\left(\dfrac{1}{99.98}+\dfrac{1}{98.97}+\dfrac{1}{97.96}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)
\(=\dfrac{1}{99}-\left(\dfrac{1}{99}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{96}+...+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{2}-1\right)\)
\(=\dfrac{1}{99}-\left(\dfrac{1}{99}-1\right)\)
\(=\dfrac{1}{99}-\dfrac{-98}{99}\)
\(=\dfrac{1}{99}+\dfrac{98}{99}\)
\(=\dfrac{99}{99}=1\)
Chúc bạn học tốt!
\(\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-...-\dfrac{1}{3.2}+\dfrac{1}{2.1}\)
=\(\dfrac{1}{99}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{97}+\dfrac{1}{96}-\dfrac{1}{96}+...+\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\)
=\(0+1\)
=\(1\)
Bạn học tốt^^
Ta có B= \(\frac{1}{2009.2010}-(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2007.2008}+\frac{1}{2008.2009}) \)
=\(\frac{1}{2009.}-\frac{1}{2010} -(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2007}-\frac{1}{2008} +\frac{1}{2008}-\frac{1}{2009}) \)
=\(\frac{1}{2009}-\frac{1}{2010}-(1-\frac{1}{2009} )\)
=\(\frac{2}{2009}-1 -\frac{1}{2010} \)
=1/50-(1-1/2+1/2-1/3+...+1/49-1/50)
=1/50-1+1/50
=1/25-1=-24/25
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{1}{50}-\dfrac{1}{50\cdot49}-\dfrac{1}{49\cdot48}-...-\dfrac{1}{2\cdot1}\)
`=`\(\dfrac{1}{50}-\left(\dfrac{1}{50\cdot49}+\dfrac{1}{49\cdot48}+...+\dfrac{1}{2\cdot1}\right)\)
`=`\(\dfrac{1}{50}-\left(\dfrac{1}{50}-\dfrac{1}{49}+\dfrac{1}{49}-\dfrac{1}{48}+...+\dfrac{1}{2}-1\right)\)
`=`\(\dfrac{1}{50}-\left(\dfrac{1}{50}-1\right)\)
`=`\(\dfrac{1}{50}-\left(-\dfrac{49}{50}\right)\)
`= 1`