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LD
18 tháng 7 2017
\(3.\)
\(\frac{x-1}{2011}+\frac{x-2}{2010}+\frac{x-3}{2009}=\frac{x-4}{2008}\)
\(\Rightarrow\)\(\frac{x-1}{2011}-1+\frac{x-2}{2010}-1+\frac{x-3}{2009}-1-\frac{x-4}{2008}+1+2=0\)
\(\Rightarrow\)\(\frac{x-1}{2011}-\frac{2011}{2011}+\frac{x-2}{2010}-\frac{2010}{2010}+\frac{x-3}{2009}-\frac{2009}{2009}-\frac{x-4}{2008}+\frac{2008}{2008}=0\)
\(\Rightarrow\)\(\frac{x-2012}{2011}+\frac{x-2012}{2010}+\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Rightarrow\)\(x-2012\left(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)=0\)
\(\Rightarrow\)\(x=2012\)
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
9 tháng 5 2018
a/\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)
=\(\frac{2^3.5^3.7^4}{2^2.5^2.7^4}\)
=2.5
=10
NA
6
NA
1
MT
11 tháng 7 2015
\(\frac{x+2}{2012}+\frac{x+3}{2011}=\frac{x+4}{2010}+\frac{x+5}{2009}\)
\(\Rightarrow\frac{x+2}{2012}+1+\frac{x+3}{2011}+1=\frac{x+4}{2010}+1+\frac{x+5}{2009}+1\)
\(\frac{x+2}{2012}+\frac{2012}{2012}+\frac{x+3}{2011}+\frac{2011}{2011}=\frac{x+4}{2010}+\frac{2010}{2010}+\frac{x+5}{2009}+\frac{2009}{2009}\)
\(\frac{x+2014}{2012}+\frac{x+2014}{2011}=\frac{x+2014}{2010}+\frac{x+2014}{2009}\)
\(\frac{x+2014}{2012}+\frac{x+2014}{2011}-\frac{x+2014}{2010}-\frac{x+2014}{2009}=0\)
\(\left(x+2014\right)\left(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)
mà \(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}\ne0\)
nên \(x+2014=0\)
\(x=-2014\)
TN
19 tháng 7 2016
a)\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)
\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{2013}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1}{2013}\)
đề sai
b)\(\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(x+2004=0\).Do \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
\(x=-2004\)
c)\(\frac{x+5}{205}-1+\frac{x+4}{204}-1+\frac{x+3}{203}-1=\frac{x+166}{366}-1+\frac{x+167}{367}-1+\frac{x+168}{368}-1\)
\(\frac{x-200}{205}+\frac{x-200}{204}+\frac{x-200}{203}=\frac{x-200}{366}+\frac{x-200}{367}+\frac{x-200}{368}\)
\(\frac{x-200}{205}+\frac{x-200}{204}+\frac{x-200}{203}-\frac{x-200}{366}-\frac{x-200}{367}-\frac{x-200}{368}=0\)
\(\left(x-200\right)\left(\frac{1}{205}+\frac{1}{204}+\frac{1}{203}-\frac{1}{366}-\frac{1}{367}-\frac{1}{368}\right)=0\)
\(x-200=0\).Do\(\frac{1}{205}+\frac{1}{204}+\frac{1}{203}-\frac{1}{366}-\frac{1}{367}-\frac{1}{368}\ne0\)
\(x=200\)
d)chịu