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Ta có : \(1-\frac{1}{2014.2013}-\frac{1}{2013.2012}-......-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2013.2014}\right)\)
\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(=1-\left(1-\frac{1}{2014}\right)\)
\(=1-1+\frac{1}{2014}\)
\(=\frac{1}{2014}\)
\(a,1-\frac{1}{2014.2013}-\frac{1}{2013.2012}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\right)\)
\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(=1-\left(1-\frac{1}{2014}\right)\)
\(=1-1+\frac{1}{2014}\)
\(=\frac{1}{2014}\)
\(\dfrac{1}{2014}-\dfrac{1}{2014.2013}-\dfrac{1}{2013.2012}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}=\dfrac{1}{2014}-\left(\dfrac{1}{2013.2014}+\dfrac{1}{2012.2013}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)=\dfrac{1}{2014}-\left(\dfrac{1}{2013}-\dfrac{1}{2014}+\dfrac{1}{2012}-\dfrac{1}{2013}+...+\dfrac{1}{2}-\dfrac{1}{3}+1-\dfrac{1}{2}\right)=\dfrac{1}{2014}-\left(1-\dfrac{1}{2014}\right)=\dfrac{1}{2014}-\dfrac{2013}{2014}=-\dfrac{1006}{1007}\)
=1/2014-(1/1*2+1/2*3+...+1/2013*2014)
=1/2014-(1-1/2+1/2-1/3+...+1/2013-1/2014)
=1/2014-1+1/2014
=1/1007-1=-1006/1007
\(\dfrac{1}{2014}-\dfrac{1}{2014.2013}-\dfrac{1}{2013.2012}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}=\dfrac{1}{2014}-\left(\dfrac{1}{2013.2014}+\dfrac{1}{2012.2013}+....+\dfrac{1}{1.2}\right)=\dfrac{1}{2014}-\left(\dfrac{1}{2013}-\dfrac{1}{2014}+\dfrac{1}{2012}-\dfrac{1}{2013}+...+1-\dfrac{1}{2}\right)=\dfrac{1}{2014}-\left(1-\dfrac{1}{2014}\right)=\dfrac{1}{2014}-\dfrac{2013}{2014}=-\dfrac{2012}{2014}=-\dfrac{1006}{1007}\)
a) \(\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{99}-\left(\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
đặt \(A=\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)
\(A=1-\frac{1}{99}\)
\(A=\frac{98}{99}\)
thay A vào, ta được :
\(\frac{1}{99}-\frac{98}{99}=\frac{-97}{99}\)
b) \(\frac{2}{100.99}-\frac{2}{99.98}-...-\frac{2}{3.2}-\frac{2}{2.1}\)
\(=\frac{2}{100.99}-\left(\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\right)\)
đặt \(A=\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\)
\(A=\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{98.99}\)
\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)
\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(A=2.\left(1-\frac{1}{99}\right)\)
\(A=2.\frac{98}{99}\)
\(A=\frac{196}{99}\)
Thay A vào, ta được :
\(\frac{2}{100.99}-\frac{196}{99}=\frac{-19598}{9900}\)
\(\left(\frac{1}{4}-x\right)\left(x+\frac{2}{5}\right)=0\)
Ta xét 2 trường hợp
\(\begin{cases}\frac{1}{4}-x=0\\x+\frac{2}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{2}{5}\end{cases}}\)
tớ mới làm bài 1 thôi bài 2 3 tớ ko có thời gian
\(A=\frac{1}{2003.2002}-\frac{1}{2002.2001}-\frac{1}{2001.2000}-....-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(\frac{1}{2003.2002}+\frac{1}{2002.2001}+\frac{1}{2001.2000}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=-\left(\frac{1}{2003}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2001}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)
\(=-\left(\frac{1}{2003}-1\right)=-\left(-\frac{2002}{2003}\right)=\frac{2002}{2003}\)
Vậy ....
\(\frac{1}{2014}-\frac{1}{2014.2013}-\frac{1}{2013.2012}-...-\frac{1}{3.2}-\frac{1}{2.1}.\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}+\frac{1}{2013.2014}\right)+\frac{1}{2014}\)
\(=\frac{1}{2014}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(=\frac{1}{2014}-1+\frac{1}{2014}=\frac{1}{1007}-1=\frac{-1006}{1007}\)
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