Đặt \(A=\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-....-\frac{1}{5.3}-\frac{1}{3.1}\)

\(\Rightarrow A=\frac{1}{99.97}-\left(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{93.95}+\frac{1}{95.97}\right)\)

\(\Rightarrow A=\frac{1}{99.97}-\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{95}-\frac{1}{97}\right)\)

\(\Rightarrow A=\frac{1}{99}-\frac{1}{97}-\frac{1}{2}\left(1-\frac{1}{97}\right)=\frac{1}{99}-\frac{1}{97}-\frac{1}{2}-\frac{1}{194}\)

k mk nha!

thanks!

nhanha!!!

Gọi A=1/99x97-1/97x95-1/95x93-...-1/5x3-1/3x1

Suy ra A=-1/1x3-1/3x5-...-1/93x95-1/95x97-1/97x99

2A=-2/1x3-2/3x5-...-1/93x95-1/95x97-1/97x99

2A=-(2/1x3+2/3x5+...+1/93x95+2/95x97+1/97x99

2A=-(1/2-1/3+1/2-1/5+...+1/93-1/95+1/95-1/97+1/97-1/99)

2A=-(1/2-1/99)

2A=-97/198

A=-97/396

Đặt A =\(\frac{1}{99x97}+\frac{1}{97x95}+...+\frac{1}{3x1}\)

2A =\(\frac{2}{99x97}+\frac{2}{97x95}+...+\frac{2}{3x1}\)

2A=\(\frac{1}{97}-\frac{1}{99}+\frac{1}{95}-\frac{1}{97}+...+\frac{1}{1}-\frac{1}{3}\)

2A=1-\(\frac{1}{99}\)=\(\frac{98}{99}\)

=> A=\(\frac{49}{99}\)

\(\frac{-92}{93}\)

EM MỚI HỌK LỚP 6 THUI ẠK CHẮC K ĐÚNG HOẶC CÓ THỂ ĐÚNG ẠK NẾU ĐÚNG THÌ K ANH NHÓE!

CHÚC ANH HỌC TỐT

`1/2 -1/6- 1/9`

`= 3/6 -1/6 -1/9`

`= 2/6 -1/9`

`= 1/3-1/9`

`=3/9-1/9`

`=2/9`

\(\dfrac{1}{2}-\dfrac{1}{6}-\dfrac{1}{9}=\dfrac{9}{18}-\dfrac{3}{18}-\dfrac{2}{18}=\dfrac{4}{18}=\dfrac{2}{9}\)

1/2 - 1/6 - 1/9 = 9/18 - 3/18 - 2/18= 4/18

Rút gọn còn 2/9

`1/( 1.2 ) + 1/( 2.3 ) + .......+1/(99.100)`

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

`=1-1/100`

`=99/100`

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

=1-1/100=99/100

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

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

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

Giúp mình

Ta có: \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2004.2005.2006}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2004.2005}-\frac{1}{2005.2006}\)

\(=\frac{1}{1.2}-\frac{1}{2005.2006}\)

\(=\frac{1}{2}-\frac{1}{4022030}\)

\(=-40220295.\)

\(=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+...-\frac{1}{2005}+\frac{1}{2004}-\frac{1}{2005}-\frac{1}{2006}\)

\(=1-\frac{1}{2006}=\frac{2005}{2006}\)