`A=1/(2xx3)+1/(3xx4)+1/(4xx5)+1/(5xx6)+1/(6xx7)+1/(7xx8)`

`=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8`

`=1/2-1/8`

`=4/8-1/8`

`=3/8`

Vậy `A=3/8`

1/2x3 +1/3x4 +1/4x5 + 1/5x6 = 7/18

\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)

\(=\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+\frac{6-5}{5\times6}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(=\frac{1}{2}-\frac{1}{6}\)

\(=\frac{1}{3}\)

A=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
A=\(\frac{1}{2}-\frac{1}{7}\)
A=\(\frac{5}{14}\)

A = 1/2 -1/3 +1/3-1/4 + 1/4-1/5 +1/5-1/6 + 1/6-1/7 =

1/2-1/7 = 5/14

1/2x3 +1/3x4 +1/4x5 +1/5x6 + 1/6x7 +1/7x8 + 1/8x9 = 7/18

= 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10 + 1/10 - 1/11 + 1/11 - 1/12

= 1/3 - 1/12 = 1/4

Sai rồi bạn ạ

Phải là 1/3 nhân 4 chứ ko phải 1/3x4

\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)

Đặt  \(A=\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}\)

\(A=\frac{3-2}{2x3}+\frac{4-3}{3x4}+\frac{5-4}{4x5}+\frac{6-5}{5x6}+\frac{7-6}{6x7}+\frac{8-7}{7x8}+\frac{9-8}{8x9}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)

\(A=\frac{1}{2}-\frac{1}{9}=\frac{9}{18}-\frac{2}{18}=\frac{7}{18}\)

1/2x3 + 1/3x4 + 1/4x5 +1/5x6 +1/6x7+1/7x8 + 1/8x9 = 7/18

= 1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/7-1/8+1/8-1/9
=1/2-1/9
= 7/18

\(B=\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)+\left(\dfrac{5}{6}+\dfrac{19}{20}+...+\dfrac{2549}{2550}\right)\)

\(B=\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+..+\dfrac{1}{50\cdot51}\right)+\left(1-\dfrac{1}{2\cdot3}\right)+\left(1-\dfrac{1}{3\cdot4}\right)+...+\left(1-\dfrac{1}{50\cdot51}\right)\)

\(B=\left(1+1+...+1\right)+\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)\)

\(B=1\cdot49=49\) (vì có (50 - 2) : 1 + 1 = 49 số hạng 1)