`A=1/(1xx2)+1/(2xx3)+1/(3xx4)+...+1/(99xx100)`

`=> A=(2-1)/(1xx2)+(3-2)/(2xx3)+...+(100-99)/(99xx100)`

`=> A=1-1/2+1/2-1/3+...+1/99-1/100`

`=> A=1-1/100`

`=> A=99/100

Sửa đề:

A = 1/(1.2) + 1/(2.3) + 1/(3.4) + ... + 1/(97.98) + 1/(98.99) + 1/(99.100)

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/97 - 1/98 + 1/98 - 1/99 + 1/99 - 1/100

= 1 - 1/100

= 99/100

muốn biết thì tra goolge bằng 260/100

\(\frac{2}{1}\cdot3\cdot\frac{2}{3}\cdot5\cdot\frac{2}{5}\cdot7\cdot\frac{2}{7}\cdot9\cdot\frac{2}{9}\cdot11\)

\(=2\cdot\left(3\cdot\frac{2}{3}\right)\cdot\left(5\cdot\frac{2}{5}\right)\cdot\left(7\cdot\frac{2}{7}\right)\cdot\left(9\cdot\frac{2}{9}\right)\cdot11\)

\(=2\cdot2\cdot2\cdot2\cdot2\cdot11\)

\(=352\)

jgtgmjhkgjbu kgug

`A=1/(1xx3)+1/(3xx5)+1/(5xx7)+...+1/(95xx97)+1/(97xx99)`

`=> 2A=2/(1xx3)+2/(3xx5)+...+2/(97xx99)`

`=> 2A=1-1/3+1/3-1/5+...+1/97-1/99`

`=> 2A=1-1/99`

`=> 2A=98/99`

`=> A=49/99`

P = \(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{98}{99}\). CMR: P \(< \frac{1}{7}\)

Đề bài đây à

\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}\)

\(2A=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{20-18}{18.19.20}=\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}=\dfrac{1}{2}-\dfrac{1}{19.20}\)

\(\Rightarrow A=\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right):2\)