1/1x2 + 1/2x3 + 1/3x4 + ... + 1/99x100 
=(1 -1/2) +(1/2 -1/3) +(1/3 - 1/4) +......+(1/99 - 1/100) 
(gạch bỏ -1/2 và 1/2 ; -1/3 và 1/3 ; .........-1/99 và 1/99) 
=1-1/100 
=99/100

tích nha

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

=1/1 - 1/100=99/100

1/1x2 + 1/2x3 +...+ 1/99x100

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

=1 - 1/100=99/100

=1-1/2+1/2-1/3+...+1/1981-1/1982

=1-1/1982

=1981/1982

Lời giải:

$\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+....+\frac{1}{1981\times 1982}$

$=\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+...+\frac{1982-1981}{1981\times 1982}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1981}-\frac{1}{1982}$

$=1-\frac{1}{1982}=\frac{1981}{1982}$

=1-1/2+1/2-1/3+...+1/1981-1/1982

=1-1/1982

=1981/1982 

A = \(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\) + \(\dfrac{1}{3\times4}\)+...+ \(\dfrac{1}{2021\times2022}\)

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+...+ \(\dfrac{1}{2021}\) - \(\dfrac{1}{2022}\)

A = 1 - \(\dfrac{1}{2022}\)

A = \(\dfrac{2021}{2022}\)

vì 1/1*2=1-1/2

   1/2*3=1/2-1/3

.....................

1/2014*2015=1/2014-1/2015

=1-1/2+1/2-1/3+1/3-....+1/2014-1/2015

=1-1/2015

=2014/2115

\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+....+\frac{1}{2014x2015}\)

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

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

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

\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{2023\times2024}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\\ =1-\dfrac{1}{2024}=\dfrac{2023}{2024}\)

1/1*2+1/2*3+...+1/2023*2024=1-1/2+1/2-1/3+...+1/2023-1/2024

=1-1/2024=2023/2024

\(...=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\)

\(=1-\dfrac{1}{2024}=\dfrac{2023}{2024}\)

Đặt A = 1/1x2 + 1/2x3 + 1/3x4 + .... + 1/99x100

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

=> A = 1 - 1/100

=> A = 99/100

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

hok

tốt

nha

1/(1×2) + 1/(2×3) + 1/(3×4) + ... + 1/(2021×2022)

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2021 - 1/2022

= 1 - 1/2022

= 2021/2022