Question

\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+.....+\frac{1}{99x100}\)

Answer 1

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

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

k cho mình nha bạn

Answer 2

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

=1-1/100=99/100

Answer 3

I don't know

Answer 4

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

= \(1-\frac{1}{100}=\frac{99}{100}\)

Answer 5

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

Answer 6

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

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

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

Answer 7

99/100 ai mik lai cho

Answer 8

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

=1/1-1/100

=100/100-1/100

=99/100