\(A=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{2}-\frac{1}{100}\)

\(\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)

\(\Rightarrow A< \frac{1}{2}\)

\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{2}-\frac{1}{100}\)

\(\Leftrightarrow A< \frac{1}{2}\)

\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{2}-\frac{1}{100}\)

\(A=\frac{49}{100}\)

Vậy A < 1/2.Vì 1/2 = 50/100 nên 49/100 > 50/100 nên A > 1/2

Ai tích cho toi thi h lai cho

Ta có: 1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100
         = 1/2-1/100
         = 50/100-1/100
         = 49/100

A = 1/2 - 1/3 + 1/3 -1/4 + 1/4 -1/5 + ...+ 1/98 - 1/99 + 1/99 - 1/100 
Ta thấy đoạn giữa sẽ bị trừ lẫn nhau nên bằng 0 
A = 1/2 - 1/100 = 49/100

tích nha

hi bn

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)

1/2.3+1/3.4+1/4.5+...+1/99.100

=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)

=\(\frac{1}{2}-\frac{1}{100}=\frac{50}{100}-\frac{1}{100}=\frac{49}{100}\)

=1/2-1/3+1/3-1/4+1/4-1/5+......+1/99-1/100

=1/2-1/100

=49/100

1/2*3+1/3*4+...+1/99*100

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

=50/100-1/100=49/100

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)\(=>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(=>\frac{1}{2}-\frac{1}{100}=>\frac{50}{100}-\frac{1}{100}=\frac{49}{100}\)

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

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

\(=\dfrac{49}{100}\)

S=1-1/100=99/100

bạn tách ra, 1/1.2=1-1/2 cứ như thế, rồi trừ đi còn 1-1/100=99/100

S = 99/100

\(\left(1+\frac{1}{2.3}\right)\left(1+\frac{1}{3.4}\right)\left(1+\frac{1}{4.5}\right)...\left(1+\frac{1}{99.100}\right)\)

\(=\left(1+\frac{1}{2}-\frac{1}{3}\right)\left(1+\frac{1}{3}-\frac{1}{4}\right)\left(1+\frac{1}{4}-\frac{1}{5}\right)...\left(1+\frac{1}{99}-\frac{1}{100}\right)\)

\(=1+\frac{1}{2}-\frac{1}{3}.1+\frac{1}{3}-\frac{1}{4}.1+\frac{1}{4}-\frac{1}{5}...1+\frac{1}{99}-\frac{1}{100}\)

\(=1+\frac{1}{2}-1.\left(\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(=1+\frac{1}{2}-1.\left(2\frac{1}{3}-2\frac{1}{4}-...-2\frac{1}{99}-\frac{1}{100}\right)\)

\(=1+\frac{1}{2}-1\left[2.\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-...-\frac{1}{99}\right)\right]-\frac{1}{100}\)

tới đây bí 

\(S=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(S=\frac{1}{2}-\frac{1}{100}\)

\(S=\frac{49}{100}\)

chúc các bạn học tốt

\(S=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(S=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(S=1\times\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(S=1\times\frac{49}{100}\)

\(S=\frac{49}{100}\)

Ta có: \(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

=>\(S=\frac{1}{2}-\frac{1}{100}=\frac{50}{100}-\frac{1}{100}=\frac{50-1}{100}=\frac{49}{100}\)

Vậy \(S=\frac{49}{100}\)