a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{7}\right)\)

\(=\frac{1}{2}.\frac{6}{7}\)

\(=\frac{3}{7}\)

b)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)

\(=\frac{1}{2}.\frac{2010}{2011}\)

\(=\frac{1005}{2011}\)

Mik nghĩ đề phải là cộng chứ

\(=\dfrac{1}{3}-\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{3}-\dfrac{1}{2}\cdot\dfrac{98}{303}=\dfrac{1}{3}-\dfrac{49}{303}=\dfrac{101-49}{303}=\dfrac{52}{303}\)

a) Đặt B= 1/1.3 + 1/3.5 + 1/5.7 + .....+ 1/19.21

Ta có: 2B= 2/1.3 + 2/3.5 + 2/5.7 + ....+ 2/19.21

= 1- 1/3 + 1/3-1/5 + 1/5-1/7 +....+ 1/19-1/21

= 1-1/21 = 20/21

=> B= 20/21 : 2 => B= 10/21

b) Như trên, ta có: 2A= 1- (1/2n + 1) => A=( 1-1/2n+1).1/2

=> A= 1/2- 1/2n+1

=> A< 1/2 ( đpcm )

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(=\frac{1}{1}-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)

\(\Rightarrow A=\frac{100}{101}:2=\frac{100}{101}\times\frac{1}{2}=\frac{50}{101}\)

\(D=\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{199.201}\)

\(D=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{199.201}\right)\)

\(D=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{199}-\frac{1}{201}\right)\)

\(D=\frac{3}{2}\left(1-\frac{1}{201}\right)\)

\(D=\frac{3}{2}.\frac{200}{201}\)

\(D=\frac{100}{67}\)

#)Giải :

\(D=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{199.201}\)

\(D=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{199.201}\right)\)

\(D=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{199}-\frac{1}{201}\right)\)

\(D=\frac{3}{2}\left(1-\frac{1}{201}\right)\)

\(D=\frac{3}{2}\times\frac{200}{201}\)

\(D=\frac{100}{67}\)

\(\frac{2}{3}D=\frac{2}{1\times3}+\cdot\cdot\cdot+\frac{2}{199\times201}\)

\(\frac{2}{3}D=1-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{199}-\frac{1}{201}\)

\(\frac{2}{3}D=1-\frac{1}{201}\)

\(\frac{2}{3}D=\frac{200}{201}\)

\(D=\frac{200}{201}:\frac{2}{3}=\frac{200\times3}{201\times2}=\frac{300}{201}\)

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{2013.2015}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{2013.2015}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2015}\right)=\frac{1}{2}.\frac{2014}{2015}=\frac{1007}{2015}\)

Vậy A=1007/2015

\(2A=2\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\right)\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\)

\(2A=1-\frac{1}{2015}\)

\(A=\frac{2014}{2015}:2\)

\(A=\frac{1007}{2015}\)

1/1.3+1/3.5+...+1/2013.2015

=1/2.(1/1-1/3+1/3-1/5+...+1/2013-1/2015)

=1/2.(1/1-1/2015)

=1/2.2014/2015

=1007/2015

A=1/1.3+1/3.5+1/5.7+...+1/2013.2015

2A=2.(1/1.3+1/3.5+1/5.7+...+1/2013.2015)

=2/1.3+2/3.5+2/5.7+...+2/2013.2015

=1-1/3+1/5-1/7+1/7-1/9+...+1/2013-1/2015

=1-1/2015

=2014/2015

=>2A=2014/2015=>A=1007/2015

A=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)

A=\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\) (phân số 1/2 là bn xem hai số của mẫu,số đuôi trừ số đầu có ra kết quả bằng tử hay ko,nếu ko bằng,bạn phải nhân thêm với 1 phân số có tử là tử của các phân số cho sẵn còn mẫu là hiệu của hai số ở mẫu)

A=\(\frac{1}{2}.\left(1-\frac{1}{2015}\right)\) (những phân số giống nhau thì cứ loại bỏ)

A=\(\frac{1}{2}.\frac{2014}{2015}=1.\frac{1007}{2015}=\frac{1007}{2015}\)

A = 1/1.3 + 1/3.5 + 1/5.7 + ... + 1/2011.2013

A = 1/2.(2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2011.2013)

A = 1/2.(1 - 1/3  + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2011 - 1/2013)

A = 1/2.(1 - 1/2013)

A = 1/2.2012/2013

A = 1006/2013

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2011.2013}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2011.2013}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\)

\(2A=1+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{2011}-\frac{1}{2011}\right)-\frac{1}{2013}\)

\(2A=1-\frac{1}{2013}\)

\(2A=\frac{2012}{2013}\)

\(A=\frac{2012}{2013}:2\)

\(A=\frac{1006}{2013}\)

~ Hok tốt ~

\(2A=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{2011\cdot2013}\right)\)

\(=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{2011\cdot2013}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\)

\(=1-\frac{1}{2013}\)

\(=\frac{2012}{2013}\)

\(A=\frac{2012}{2013}\div2=\frac{2012}{2013\cdot2}\)

                                                                     \(#Louis\)

\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2009\cdot2011}\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2009\cdot2011}\right)\)

\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)

\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{2011}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2010}{2011}=\dfrac{1005}{2011}\)

= 1/2 . (1/1 - 1/3 + 1/3 - 1/5 +... + 1/2009 - 1/2011)

= 1/2 . (1/1 - 1/2011)

= 1/2 . 2010 / 2011

= 1005/2011