Mình ko chép đề nx nha

A = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{999}-\frac{1}{1000}\)

A =  \(\frac{1}{1}-\frac{1}{1000}\)

A = \(\frac{1000}{1000}-\frac{1}{1000}=\frac{999}{1000}\)

B = \(\frac{1}{501}-\frac{1}{1000}+\frac{1}{502}-\frac{1}{999}+...\frac{1}{1}+...+\frac{1}{999}-\frac{1}{502}+\frac{1}{1000}+\frac{1}{501}\)

B = \(\frac{1}{501}-\frac{1}{501}+\frac{1}{1000}-\frac{1}{1000}+\frac{1}{502}-\frac{1}{502}+\frac{1}{999}-\frac{1}{999}+...+\frac{1}{1}\)

B = \(\frac{1}{1}=1\)

Vậy \(\frac{A}{B}=\frac{\frac{999}{1000}}{1}=\frac{999}{1000}\)

Thx Bn nhiều <3

Cho mk hỏi cái \(\frac{1}{2}-\frac{1}{3}\)ở dâu thế bn?

a.

$A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{1000-999}{999.1000}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}$

$=1-\frac{1}{1000}=\frac{999}{1000}$

b.

$5B=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+....+\frac{5}{495.500}$

$=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+....+\frac{500-495}{495.500}$

$=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{495}-\frac{1}{500}$

$=1-\frac{1}{500}=\frac{499}{500}$

$\Rightarrow B=\frac{499}{500}: 5= \frac{499}{2500}$

c.

$2C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{998.999.100}$

$=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{1000-998}{998.999.1000}$

$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{998.999}-\frac{1}{999.1000}$

$=\frac{1}{1.2}-\frac{1}{999.1000}=\frac{499499}{999000}$

$\Rightarrow C=\frac{499499}{999000}:2=\frac{499499}{1998000}$

999/1000(hình như v)

Áp dụng công thức \(\dfrac{1}{k\left(k+1\right)}=\dfrac{1}{k}-\dfrac{1}{k+1}\), ta có:

\(A=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{999}-\dfrac{1}{1000}\right)=1-\dfrac{1}{1000}=\dfrac{999}{1000}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}\)

\(A=1-\dfrac{1}{1000}\)

\(A=\dfrac{999}{1000}\)

a, 1/1.2+1/2.3+1/3.4+...+1/999.1000

=  1/1-1/2+1/2-1/3+1/3-1/4+....+1/999-1/1000

=   1/1-1/1000

=   999/1000

b, 1/2.4+1/4.6+1/6.8+1/8.10

=  1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10

=  1/2-1/10

=   4/10  =2/5

bài 2:

a)\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)

\(=1-\frac{1}{1000}\)

\(=\frac{999}{1000}\)

mk ko biết bn có sai đề ko nhưng mk chỉ lm theo ý mk hiểu thôi! sai thì thôi nha!

bn làm như vầy nè

a=1/51+1/52+...+1/100

A=1/3.1/7 + 1/2.1/26+....1/2.1/50

A=1/3-1/7+1/2-1/26+...1/2-1/50

A=1/3-1/50

A=47/50

như vầy đó bn tin mik đi