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NT
1
11 tháng 5 2015
\(\frac{1}{1}+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x.\left(x+1\right)}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=2.\left(1-\frac{1}{x+1}\right)\)
\(=2-\frac{2}{x+1}\) mà \(\frac{1}{1}+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=1\frac{2008}{2010}\)
=> \(2-\frac{2}{x+1}=1\frac{2008}{2010}=>\frac{2}{x+1}=\frac{2}{2010}=>x+1=2010=>x=2009\)
đúng cái nhé
WJ
0
LD
18 tháng 7 2017
\(3.\)
\(\frac{x-1}{2011}+\frac{x-2}{2010}+\frac{x-3}{2009}=\frac{x-4}{2008}\)
\(\Rightarrow\)\(\frac{x-1}{2011}-1+\frac{x-2}{2010}-1+\frac{x-3}{2009}-1-\frac{x-4}{2008}+1+2=0\)
\(\Rightarrow\)\(\frac{x-1}{2011}-\frac{2011}{2011}+\frac{x-2}{2010}-\frac{2010}{2010}+\frac{x-3}{2009}-\frac{2009}{2009}-\frac{x-4}{2008}+\frac{2008}{2008}=0\)
\(\Rightarrow\)\(\frac{x-2012}{2011}+\frac{x-2012}{2010}+\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Rightarrow\)\(x-2012\left(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)=0\)
\(\Rightarrow\)\(x=2012\)
NN
2
HC
23 tháng 2 2017
\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}=1\frac{2008}{2010}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=1\frac{2008}{2010}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=1\frac{2008}{2010}\):2
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2010}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2009}{2010}\)
\(\Rightarrow1-\frac{2009}{2010}=\frac{1}{x+1}\)
\(\Rightarrow\frac{1}{2010}=\frac{1}{x+1}\)
\(\Rightarrow x=2009\)
nha !
MT
23 tháng 2 2017
Ta có :A=1+\(\frac{2}{6}\)+\(\frac{2}{12}\)+......+\(\frac{2}{x\left(x+1\right)}\)=\(\frac{4018}{2010}\)
\(\Rightarrow\)A=\(\frac{2}{2.3}\)+\(\frac{2}{3.4}\)+...+\(\frac{2}{x\left(x+1\right)}\)=\(\frac{2008}{2010}\)
\(\Rightarrow\)A=2(\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{x\left(x+1\right)}\))=\(\frac{2008}{2010}\)
\(\Rightarrow\)A=2(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...+\(\frac{1}{x}\)-\(\frac{1}{x+1}\))=\(\frac{2008}{2010}\)
\(\Rightarrow\)A=2(\(\frac{1}{2}\)-\(\frac{1}{x+1}\))=\(\frac{2008}{2010}\)
\(\Rightarrow\)A=\(\frac{1}{2}\)-\(\frac{1}{x+1}\)=\(\frac{502}{1005}\)
\(\Rightarrow\)\(\frac{1}{x+1}\)=\(\frac{1}{2010}\)\(\Rightarrow\)x+1=2010\(\Rightarrow\)x=2009
21 tháng 3 2017
a)\(\frac{5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{4}-7x\)
\(\Leftrightarrow\frac{5}{2}-1+x=\frac{1}{4}-7x\)
\(\Leftrightarrow8x=-\frac{5}{4}\)
\(\Leftrightarrow x=-\frac{5}{32}\)
c)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2003}\)
\(\Leftrightarrow x+1=2003\)
\(\Leftrightarrow x=2002\)
28 tháng 4 2017
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-
S
28 tháng 4 2017
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
Lời giải:
$\frac{1}{1}+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x(x+1)}=1\frac{2008}{2010}$
$\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x(x+1)}=\frac{2009}{1005}$
$2(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x(x+1)})=\frac{2009}{1005}$
$2(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1})=\frac{2009}{1005}$
$2(1-\frac{1}{x+1})=\frac{2009}{1005}$
$\frac{2x}{x+1}=\frac{2009}{1005}$
$\Rightarrow 2009(x+1)=2010x$
$\Rightarrow x=2009$