\(3.\)

\(\frac{x-1}{2011}+\frac{x-2}{2010}+\frac{x-3}{2009}=\frac{x-4}{2008}\)

\(\Rightarrow\)\(\frac{x-1}{2011}-1+\frac{x-2}{2010}-1+\frac{x-3}{2009}-1-\frac{x-4}{2008}+1+2=0\)

\(\Rightarrow\)\(\frac{x-1}{2011}-\frac{2011}{2011}+\frac{x-2}{2010}-\frac{2010}{2010}+\frac{x-3}{2009}-\frac{2009}{2009}-\frac{x-4}{2008}+\frac{2008}{2008}=0\)

\(\Rightarrow\)\(\frac{x-2012}{2011}+\frac{x-2012}{2010}+\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)

\(\Rightarrow\)\(x-2012\left(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)=0\)

\(\Rightarrow\)\(x=2012\)

Bài 2:b)Ta có:

D=(51*52*53*...*100):2^50.

=(51*53*55*...*99)*(52*54*56*...*100):2^50.

Khử 51*53*55*...*99 thì cần so sánh 1*3*5*...*41 với (52*54*56*...*100):2^50.

Lại có:

52*54*56*...*100:2^50=(52:2)*(54:2)*...*(100:2):(2^25)   (vì 52;54;56;...;100 có 25 thừa số.

=26*27*28*...*50:2^25.

=(27*29*31*...*49)*(26*28*30*...*50):2^25

Khử với 1*3*5*...*49 thì cần so sánh 1*3*5*...*25 với (26*28*30*...*50):2^25.

Lại có:

26*28*30*...*50:2^25=(26:2)*(28:2)*(30:2)*...*(50:2):2^12(vì 26;28;30;...;50 có 13 thừa số).

=13*14*15*...*25:2^12.

=(13*15*17*19*21*23*25)*(14*16*18*20*22*24):2^12.

Khử với 1*3*5*...*25 thì cần so sánh 1*3*5*7*9*11 với (14*16*18*20*22*24):2^12.

Giờ số nhỏ rồi bấm máy tính so sánh là được.\

=>C=D.

Vậy C=D.

mấy câu kia dễ rồi tự l;àm nha mk nhắc câu khó thôi.

tk cho mk nha các bn.

-chúc ai tk mk học giỏi-

1/

a, x + (x+1) + (x+2) +...+ (x+100) = 2029099

(x+x+x+...+x) + (1+2+...+100) = 2029099

2011x + 2021055 = 2029099

2011x = 2029099 - 2021055 

2011x = 8044

x = 8044 : 2011

x = 4

b, 2+4+6+....+2x = 210

=> 2(1+2+3+...+x) = 210

=> \(\frac{2x\left(x+1\right)}{2}=210\)

=> x(x+1) = 14.15

=> x = 14

2/

a, Vì B < 1

\(\Rightarrow B< \frac{2009^{2009}+1+2008}{2009^{2010}+1+2008}=\frac{2009^{2009}+2009}{2009^{2010}+2009}=\frac{2009\left(2009^{2008}+1\right)}{2009\left(2009^{2009}+1\right)}=\frac{2009^{2008}+1}{2009^{2009}+1}\)= A

Vậy A > B

b, Ta có:

\(D=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}.....\frac{100}{2}=\frac{51.52.53....100}{2^{50}}\)

\(=\frac{\left(51.52.53....100\right)\left(1.2.3.4....50\right)}{2^{50}.\left(1.2.3.4....50\right)}\)

\(=\frac{1.2.3.4.5.6.....100}{\left(2.1\right)\left(2.2\right).\left(2.3\right).....\left(2.50\right)}\)

\(=\frac{1.2.3.4.5.6......100}{2.4.6........100}=\frac{\left(1.3.5....99\right)\left(2.4.6....100\right)}{2.4.6....100}\)

\(=1.3.5....99=C\)

Vậy C = D

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\frac{x\left(x+2\right)}{2}}=1\frac{2009}{2011}\)

\(\Leftrightarrow1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{1}{x\left(x+2\right)}=1\frac{2009}{2011}\)

\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+2\right)}=1\frac{2009}{2011}-1\)

\(\Leftrightarrow\left[2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)\right]=\frac{2009}{2011}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+2}\right)=\frac{2009}{2011}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+2}=\frac{2009}{2011}\div2=\frac{2009}{4022}\)

\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{2}-\frac{2009}{4022}=\frac{1}{2011}\)

\(\Leftrightarrow x=2011-2=2009\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)

\(\Rightarrow2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}\div2\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2007}{4018}\)

\(\Rightarrow\frac{1}{x+1}=\frac{2}{4018}=\frac{1}{2009}\)

\(\Rightarrow x+1=2009\)

\(\Rightarrow x=2008\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

=>\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2007}{4018}\)(nhân cả hai vế với \(\frac{1}{2}\))

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)= \(\frac{2007}{4018}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2007}{4018}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)

\(\frac{1}{x+1}\)=\(\frac{1}{2}-\frac{2007}{4018}\)

\(\frac{1}{x+1}=\frac{1}{2009}\)

x+1=2009

x=2009-1=2008

Vậy x bằng 2008

\(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2009}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2009}\)

\(\Rightarrow2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2009}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2009}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2003}{2009}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{2009}\div2\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4018}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4018}\)

\(\Rightarrow\frac{1}{x+1}=\frac{3}{2009}\)

\(\Rightarrow\frac{3}{3\left(x+1\right)}=\frac{3}{2009}\)

\(\Rightarrow3\left(x+1\right)=2009\)

\(\Rightarrow3x+3=2009\)

\(\Rightarrow3x=2006\)

\(\Rightarrow x=\frac{2006}{3}\)

2.[1/6+1/12+1/20+...+1/x.(x+1)]=2009/2011

2.[1/2.3+1/3.4+1/4.5+...+1/x(x+1)]=2009/2011

1/2-1/3+1/3-1/4+...+1/x-1/(x+1)=2009/4022

1/2-1/(x+1)=2009/4022

1/(x+1)=1/2001

x+1=2011

x=2010

\(=>\frac{2}{3.2}+\frac{2}{6.2}+\frac{2}{10.2}+...+\frac{2}{x.\left(x+1\right):2.2}=\frac{2009}{2011}\)

\(=>\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=\frac{2009}{2011}\)

\(=>2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(=>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2009}{2011}:2\)

\(=>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{4022}\)

\(=>1-\frac{1}{x+1}=\frac{2009}{4022}\)

\(=>\frac{1}{x+1}=1-\frac{2009}{4022}\)

\(=>\frac{1}{x+1}=\frac{2013}{4022}\)

\(=>\frac{2013}{2013.\left(x+1\right)}=\frac{2013}{4022}\)

=>2013.(x+1)=4022

=>x+1=4022/2013

=>x=4022/2013-1

=>x=2009/2013

de thieu

x=-2004

\(\frac{x+2}{2012}+\frac{x+3}{2011}=\frac{x+4}{2010}+\frac{x+5}{2009}\)

\(\Rightarrow\frac{x+2}{2012}+1+\frac{x+3}{2011}+1=\frac{x+4}{2010}+1+\frac{x+5}{2009}+1\)

\(\frac{x+2}{2012}+\frac{2012}{2012}+\frac{x+3}{2011}+\frac{2011}{2011}=\frac{x+4}{2010}+\frac{2010}{2010}+\frac{x+5}{2009}+\frac{2009}{2009}\)

\(\frac{x+2014}{2012}+\frac{x+2014}{2011}=\frac{x+2014}{2010}+\frac{x+2014}{2009}\)

\(\frac{x+2014}{2012}+\frac{x+2014}{2011}-\frac{x+2014}{2010}-\frac{x+2014}{2009}=0\)

\(\left(x+2014\right)\left(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)

mà \(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}\ne0\)

nên \(x+2014=0\)

      \(x=-2014\)