đặt A=1/1x2+1/2x3+1/3x4+1/4x5+1/5x6

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}\)

\(=1-\frac{1}{6}\)

\(=\frac{5}{6}\)

 có k không

= 1/1-1/2+1/2-1/3+1/3-...-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:

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

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

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

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

chịu thua

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