\(=\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\) 

\(=\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\) 

\(=\frac{1}{99}-\frac{98}{99}=-\frac{97}{99}\)

D =1/99 -1/99.98-1/98.97-...-1/3.2-1/2.1
=1/99-(1/99.98+1/98.97-...-1/3.2+1/2.1)
=1/99-(1/1.2+1/2.3+1/3.4+...+1/98.99)
=1/99-(1/1-1/2+1/2-1/3+1/3-1/4+1/4-...+1/98-1/99)
=1/99-(1/1-1/99)
=1/99-98/99
=-97/99

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\(D=\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(D=\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

\(D=\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)

\(D=\frac{1}{99}-\frac{98}{99}=-\frac{97}{99}\)

=1

Đặt \(A=\frac{1}{99}-\frac{1}{99\cdot98}-\frac{1}{98\cdot97}-\frac{1}{97\cdot96}-\cdots-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)

\(A=\frac{1}{99}-\left(\frac{1}{98}-\frac{1}{99}\right)-\left(\frac{1}{97}-\frac{1}{98}\right)-\left(\frac{1}{96}-\frac{1}{97}\right)-\cdots-\left(\frac12-\frac13\right)-\left(\frac11-\frac12\right)\)

\(A=\frac{1}{99}-\frac{1}{98}+\frac{1}{99}-\frac{1}{97}+\frac{1}{98}-\frac{1}{96}+\frac{1}{97}-\cdots-\frac12+\frac13-1+\frac12\)

\(A=\left(\frac{1}{99}-\frac{1}{99}\right)+\left(\frac{1}{98}-\frac{1}{98}\right)+\left(\frac{1}{97}-\frac{1}{97}\right)+\left(\frac{1}{96}-\frac{1}{96}\right)+\cdots+\left(\frac12-\frac12\right)-1\)

\(A=0+0+0+0+\cdots+0+\left(-1\right)\)

\(A=-1\)

Vậy A = -1

\(P=\frac{1}{99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{98.99}\right)\)

\(=\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)

\(=\frac{1}{99}-\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{99}-\frac{98}{99}\)

\(=-\frac{97}{99}\)

Vậy \(P=-\frac{97}{99}\)

P=-1/1.2-1/2.3-...-1/98.99-1/99

P=-(1/1.2+1/2.3+...+1/98.99+1/99)

P=-1

\(P=\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-\frac{1}{97.96}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{99}-\left(\frac{1}{99}-\frac{1}{98}\right)-\left(\frac{1}{98}-\frac{1}{97}\right)-\left(\frac{1}{97}-\frac{1}{96}\right)-...-\left(\frac{1}{3}-\frac{1}{2}\right)-\frac{1}{2}\)

\(=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+\frac{1}{97}-\frac{1}{97}+\frac{1}{96}-...-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}\)

\(=0\)

ĐS: \(0\)

=\(\frac{1}{99}\)-\(\frac{1}{99}\)-\(\frac{1}{98}\)-\(\frac{1}{98}\)-.................-\(\frac{1}{3}\)-\(\frac{1}{2}\)-\(\frac{1}{2}\)-1

=\(\frac{1}{99}\)-(\(\frac{1}{99}\)+\(\frac{1}{98}\)+..............+\(\frac{1}{3}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)+1)

=\(\frac{1}{99}\)-......

hình như sai rùi????

bn kurokawa sai rồi

E=1/99-(1/99.98+1/98.97+....+1/2.1)

E=1/99-(1/1-1/2+1/2-1/3+....+1/98-1/99)

E=1/99-(1-1/99)

E=1/99-98/99

E=-97/99

\(\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{2.1}\right)\)

\(\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)

\(\frac{1}{99}-\left(1-\frac{1}{99}\right)\)

\(\frac{1}{99}-\frac{98}{99}=-\frac{97}{99}\)

Đúq nhaaa

= (1/99-1/100)- (1/98-1/99)-...(1/1-1/2)

= -(1/1-1/2+1/3-1/4+...+1/99-1/100)

=-(1/1-1/100)

=-99/100

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