\(\frac{2011x2010-1}{2009x2011+2010}=\frac{2011x\left(2009+1\right)-1}{2009x2011+2010}\)

\(=\frac{2011x2009+2011-1}{2009x2011+2010}=\frac{2011x2009+2010}{2009x2011+2010}=1\)

b.  Gọi tổng trên là A

=> A=10,11+11,12+12,13+....+97,98+98,99+99,100

=> A-99,1=10,11+11,12+12,13+.....+97,98+98,99

SSH của A-99,1 là

(98,99-10,11):1,01+1=89(SH)

Giá trị của A-99,1 là

\(\frac{89}{2}.\left(10,11+98,99\right)=4854,95\)

Vì A-99.1=4854,95

=> A=4854,95+99,1

=>A=4954.05

****

a) \(\frac{2011.2010-1}{2009.2011+2010}=\frac{2011.2009+2011-1}{2009.2011+2010}=\frac{2011.2009+2010}{2009.2011+2010}=1\)

. là nhân nha

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.

\(\text {#DNamNgV}\)

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

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

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

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

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

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

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

 A = 1/1x3 + 1/3x5 + 1/5x7 +.........+ 1/2009x2011

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

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

    = 1/1 - 1/2011

    = 2010/2011

Lời giải:

$t=\frac{2009\times 2010+2000}{2011\times 2010-2020}$

$=\frac{2010\times (2011-2)+2000}{2011\times 2010-2020}$

$=\frac{2010\times 2011-2\times 2010+2000}{2011\times 2010-2020}$

$=\frac{2010\times 2011-2020}{2011\times 2010-2020}=1$

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

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

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

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

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

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

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

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

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