\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{1}{k}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Leftrightarrow\frac{1}{2}=\frac{1}{k}\Rightarrow k=2\)

k=2

chuan 100%

\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{n+2-n}{n\left(n+1\right)\left(n+2\right)}=\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{n\left(n+2\right)}\)

\(\Rightarrow\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

\(=\frac{1}{1.2}-\frac{1}{99.100}\)

\(\Rightarrow\frac{1}{1.2.3}+...+\frac{1}{98.99.100}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Rightarrow k=2\)

A=1/100

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

\(A=\frac{99}{100}\)

1/1.2 + 1/2.3 + 1/3.4 +......+1/99.100

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

= [ ( -1/2 +1/2) +( -1/3+1/3) + (-1/4 + 1/4) +..... +( -1/99+1/99 ) ] + ( 1/1 + -1/100 ) 

= 0 + 99/100

= 99/100

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

\(=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)

Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

\(A=\frac{99}{100}< 1\) ( điều phải chứng minh ) 

Vậy \(A< 1\)

Chúc bạn học tốt ~ 

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

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

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

\(=\frac{99}{100}< 1\left(\text{đ}pcm\right)\)

vậy:\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}< 1\)

k mk bạn nha:)

Ta có : \(S=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{99.100}\right)\)

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

\(=2.\frac{99}{100}=\frac{99}{50}\)

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

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

S= 2.\(\frac{99}{100}\)

S=\(\frac{99}{50}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

Xét thấy: \(\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2};\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3};.....\)

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

A=1/1.2+1/2.3+1/3.4+...+1/99.100

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

=1/1-1/100=100/100-1/100=99/100

 Vậy A=99/100

Ta có : \(A=2+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)

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

\(\Rightarrow A=2+\left(1-\frac{1}{100}\right)\)

\(\Rightarrow A=2+\frac{99}{100}=\frac{299}{100}\)

Ta có : A=\(2+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

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

\(\Rightarrow A=2+\left(1-\frac{1}{100}\right)\)

\(\Rightarrow A=2+\frac{99}{100}\)

\(\Rightarrow A=\frac{299}{100}\)

Can you k for me,Natsu drangeel!