\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}\)

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

=\(1-\dfrac{1}{5}\)

=\(\dfrac{4}{5}\)

1/1 - 1/2+ 1/2 - 1/3 + 1/3 - 1/4+ 1/4 - 1/5

 = 1/1 - 1/5

= 4/5

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

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(A=\dfrac{1}{1}-\dfrac{1}{50}\)

\(A=\dfrac{49}{50}\)

A = 49/50

A = 1/1.2 +1/2.3 +1/3.4 +...+1/49.50    
A = 1 +1/2 -1/2+1/3-1/3+1/4-...-1/49 +1/50    

A = 1 - 1/50   
A=49/50

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:

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

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

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

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

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

= 1/1 - 1/100

= 99/100

Học từ lớp 4 rồi :V

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.

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

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

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

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

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

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{n\left(n+1\right)}\)

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

= 1 - \(\dfrac{1}{n+1}\) = \(\dfrac{n}{n+1}\)

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{19\cdot20}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

\(=1-\dfrac{1}{20}=\dfrac{19}{20}\)

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+....+\dfrac{1}{19\cdot20}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{20}\)

\(A=1-\dfrac{1}{20}\)

\(A=\dfrac{19}{20}\)

a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\)

\(2A=2\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{101}}\)

\(2A-A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{100}}\)

\(A=1-\dfrac{1}{2^{100}}\)

b) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2023\cdot2024}\)

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

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

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

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

cứu 

okee laaa