Giải:

\(\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(=-\left(-\dfrac{1}{99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+\dfrac{1}{97.96}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)

\(=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}-\dfrac{1}{99}\right)\)

\(=-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}-\dfrac{1}{99}\right)\)

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

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

Vậy ...

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

=-2/99+1=97/99

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

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

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

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

=1 nha bn, chắc vậy

mình hỏi câu này chứng minh 1+1=2

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

\(=\frac{1}{99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\right)\)

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

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

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

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

Ủng hộ ! 

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

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

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

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

Bạn tham khảo :

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#Duongw

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

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

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

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

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

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

\(\frac{1}{99}+1=\frac{100}{99}\)

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

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

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

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

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

cho mk sửa dòng thứ hai:

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

Từ đó bn làm tiếp nha

sorry mn