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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 :
\(\frac{2}{100.99}-\frac{196}{99}=\frac{-19598}{9900}\)
\(F=-\dfrac{1}{1.2}-\dfrac{1}{2.3}-...-\dfrac{1}{2014.2015}-\dfrac{1}{2015.2016}\)
\(\Rightarrow-F=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}+\dfrac{1}{2015}-\dfrac{1}{2016}=1-\dfrac{1}{2016}=\dfrac{2015}{2016}\)\(\Rightarrow F=\dfrac{-2015}{2016}\)
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 ...
Ta có:
\(P=\frac{1}{2000.1999}-\frac{1}{1999.1998}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow P=\frac{1}{1999.2000}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}\right)\)
\(\Rightarrow P=\frac{1}{1999.2000}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\right)\)
\(\Rightarrow P=\frac{1}{1999.2000}-\left(1-\frac{1}{1999}\right)\)
\(\Rightarrow P=\frac{1}{1999.2000}-\frac{1998}{1999}\)
\(\Rightarrow P=\frac{1}{1999}-\frac{1}{2000}-\frac{1998}{1999}\)
\(\Rightarrow P=\left(\frac{1}{1999}-\frac{1998}{1999}\right)-\frac{1}{2000}\)
\(\Rightarrow P=\frac{-1997}{1999}-\frac{1}{2000}\)
\(\Rightarrow P+\frac{1997}{1999}=\frac{-1997}{1999}-\frac{1}{2000}+\frac{1}{1997}\)
\(\Rightarrow P+\frac{1997}{1999}=\frac{-1}{2000}\)
Vậy....
\(\Rightarrow P=\frac{1}{2000.1999}-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{1998.1999}\right)\)
\(=\frac{1}{2000.1999}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\right)\)
\(=\frac{1}{2000.1999}-\left(1-\frac{1}{1999}\right)\)
\(=\frac{1}{1999.2000}-\frac{1998}{1999}\)
\(\Rightarrow P+\frac{1997}{1999}=\frac{1}{1999.2000}-\frac{1998}{1999}+\frac{1997}{1999}\)
\(=\frac{-1}{2000}\)
P= \(\frac{1}{2000.1999}\)- (\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}\))
= \(\frac{1}{1999}-\frac{1}{2000}\)- (\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\))
= \(\frac{1}{1999}-\frac{1}{2000}\)- ( \(1-\frac{1}{1999}\))
= \(\frac{1}{1999}-\frac{1}{2000}-\frac{1998}{1999}\)
= \(\frac{-1997}{1999}-\frac{1}{2000}\)
=) P + \(\frac{1997}{1999}\)= \(\frac{-1997}{1999}-\frac{1}{2000}+\frac{1997}{1999}=\frac{-1}{2000}\)
Gọi \(A=\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow A=\frac{1}{99.100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)
\(\Rightarrow A=\frac{1}{9900}-\left(1-\frac{1}{99}\right)\)
\(\Rightarrow A=\frac{1}{9900}-\frac{98}{99}=\frac{1}{9900}-\frac{9800}{9900}\)
\(\Rightarrow A=\frac{-9799}{9900}\)
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}-\frac{1}{2.1}=-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)=-\left(1-\frac{1}{100}\right)=-\frac{99}{100}\)