\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)

\(=\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+....+\frac{100}{99.100}-\frac{99}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

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

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

1/1.2+1/2.3+1/3.4+.......=1/99.100

A=1-1/2+1/2-1/3+....+1/99-1/100

A=1-1/100

A=99/100

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

= 1/1 - 1/100

= 99/100

Học từ lớp 4 rồi :V

= 1 -1/2 + 1/2 - 1/3 +......+1/99 - 1/100

= 1 -1/100

= 99/100

***Ai k mk mk k lại !!***

1/1*2+1/2*3+1/3*4+...+1/99*100

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

=1-1/100

=99/100

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}\)

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

=1-1/100

=99/100

=1−1/2+1/2−1/3+1/3−1/4+...+1/99−1/100
=1 − 1/100 = 99/100

1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100

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

= 1 - 1/100

= 99/100

\(P=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(P=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(P=1-\frac{1}{100}\)

\(P=\frac{99}{100}\)

P = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4+ 1/4 - 1/5 + ............. + 1/99 - 1/100

   = 1/1 - 1/100

   = 99/100

câu dễ lắm các bạn ạ

A=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{99}-\dfrac{1}{100}\)

=\(\dfrac{1}{1}-\dfrac{1}{100}\)

=\(\dfrac{99}{100}\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=1-\dfrac{1}{100}=\dfrac{99}{100}\)

\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{99}{100}\)

\(P=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{99.100}\)

\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(P=1+\left(\dfrac{-1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{-1}{3}+\dfrac{1}{3}\right)+..+\left(\dfrac{-1}{99}+\dfrac{1}{99}\right)+\dfrac{-1}{100}\)

\(P=1+0+0+....+0+\dfrac{-1}{100}\)

\(P=1+\dfrac{-1}{100}\)

\(P=\dfrac{99}{100}\)

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

=1-1/100

=99/100

hi bn

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)