1/1*2 + 1/2*3 + 1/3*4 + ... + 1/2013*2014 + 1/2014*2015

= 1 -1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2013 - 1/2014 + 1/2014 - 1/2015

=1-1/2015

=2014/2015

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

=1-1/2015

=2014/2015

đây là toán lớp 5 mà

Gọi biểu thức trên là A, ta có :

A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 2014x2015

A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 2014x2015x3

A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 2014x2015x(2016-2013)

A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 2014x2015x2016 - 2014x2015x2013.

A x 3 = 2014x2015x2016

A = 2014x2015x2016 : 3

A = 2727117120

=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 = \(\dfrac{2021}{2022}\)

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