2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2009.2011

= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2009 - 1/2011

= 1 - 1/2011

= 2010/2011

A = \(\dfrac{1}{3\times5}\) + \(\dfrac{1}{5\times7}\) + \(\dfrac{1}{7\times9}\)+...+ \(\dfrac{1}{2009\times2011}\)

A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{2}{3\times5}\) + \(\dfrac{2}{5\times7}\)+ \(\dfrac{2}{7\times9}\)+...+ \(\dfrac{1}{2009\times2011}\))

A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\)+...+ \(\dfrac{1}{2009}\) - \(\dfrac{1}{2011}\))

A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{2011}\))

A =  \(\dfrac{1}{2}\) \(\times\)  \(\dfrac{2008}{6033}\)

A = \(\dfrac{1004}{6033}\)

\(\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{2}{7\times9}+..+\dfrac{1}{2009\times2011}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\\ =\dfrac{1}{3}-\dfrac{1}{2011}\)

Đến đây bn tự tính nhé.

\(\text {#DNamNgV}\)

\(\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+...+\dfrac{1}{2009\times2011}\)

\(=\dfrac{1}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{2009\times2011}\right)\)

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

\(=\dfrac{1}{2}\times\left[\dfrac{1}{3}-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{2011}\right]\)

\(=\dfrac{1}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{2011}\right)\)

\(=\dfrac{1}{2}\times\dfrac{2008}{6033}\)

\(=\dfrac{1004}{6033}\)

Tớ không chép lại đề nữa nhé:

=\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{2009.2011}\right)\)=\(\frac{1}{2}.\left(\frac{3-1}{1-3}+\frac{7-5}{5-7}+...+\frac{2011-2009}{2009-2011}\right)\)

= \(\frac{1}{2}.\left(\frac{3}{1.3}-\frac{1}{1.3}+\frac{5}{3.5}-\frac{3}{3.5}+...+\frac{2011}{2009.2011}-\frac{2009}{2009.2011}\right)\)

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

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

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

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

bạn ơi đó là dấu nhân hay chữ ''x'' vậy?

nếu là dấu nhân thì như sau:

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{2009\cdot2011}=1\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{5}+...+\frac{1}{2009}\cdot\frac{1}{2011}=1\cdot\frac{1}{2011}=\frac{1}{2011}\)

CHẮC LÀ ĐÚNG ĐÓ BẠN. NHỚ K VÀ KB VS MK NHA. CHÚC BẠN HỌC TỐT. -_-

a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)

Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)

\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)

\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)

\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)

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

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

\(\dfrac{1}{x+1}=\dfrac{1}{100}\)

\(\Rightarrow x+1=100\)

\(x=100-1\)

\(x=99\)

câu b thiếu kết quả đúng không bn?

Công thức: \(\dfrac{1}{a\times b}=\) 1/ khoảng cách giữa a và b \(\times\left(\dfrac{1}{a}-\dfrac{1}{b}\right)\)

* Bạn làm theo công thức và vẫn dụng câu b nhé.

\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.+\frac{1}{101}-\frac{1}{103}\right)\)

\(\frac{1}{2}\left(1-\frac{1}{103}\right)=\frac{1}{2}\cdot\frac{100}{103}=\frac{50}{103}\)

xong r đó

Ta có:

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{101.103}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{103}\right)=\frac{50}{103}\)

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{101\cdot103}\)

\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{101}-\frac{1}{103}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{103}\right)\)

\(=\frac{1}{2}\cdot\frac{102}{103}=\frac{102}{206}\)

\(=\frac{51}{103}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2009.2011}\)

\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2009.2011}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2011}\right)=\frac{1}{2}.\frac{2008}{6033}=\frac{1004}{6033}\)

\(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+.....+\frac{1}{2009x2011}\)

\(=\frac{1.2}{3.5.2}+\frac{1.2}{5.7.2}+\frac{1.2}{7.9.2}+....+\frac{1.2}{2009.2011.2}\)

\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{2009.2011}\right)\)

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

\(=\frac{1}{2}.\frac{2008}{6033}=\frac{2008}{12066}\)

=(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9) chia 2 

=(1-1/9)chia 2

=8/9 chia 2

=4/9

Đặt �=11�3+13�5+15�7+17�9A=1x31​+3x51​+5x71​+7x91​

2�=21�3+23�5+25�7+27�92A=1x32​+3x52​+5x72​+7x92​

2�=11−13+13−15+...+17−192A=11​−31​+31​−51​+...+71​−91​

2�=11−19=892A=11​−91​=98​

�=89.12=49A=98​.21​=94​
 

a, Đặt :

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+..............+\dfrac{1}{19.21}\)

\(\Leftrightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+............+\dfrac{2}{19.21}\)

\(\Leftrightarrow2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+..........+\dfrac{1}{19}-\dfrac{1}{21}\)

\(\Leftrightarrow2A=1-\dfrac{1}{21}\)

\(\Leftrightarrow2A=\dfrac{20}{21}\)

\(\Leftrightarrow A=\dfrac{10}{21}\)

b, \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...........+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)

\(\Leftrightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+............+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)

\(\Leftrightarrow2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\)

\(\Leftrightarrow2A=1-\dfrac{1}{2n+1}\)

\(\Leftrightarrow2A=\dfrac{2n}{2n+1}\)

\(\Leftrightarrow A=\dfrac{n}{2n+1}\)