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\(C=\frac{4^1-1}{4^1}+\frac{4^2-1}{4^2}+...+\frac{4^{2009}-1}{4^{2009}}+\frac{4^{2010}-1}{4^{2010}}\)
\(C=\frac{4^1}{4^1}-\frac{1}{4^1}+\frac{4^2}{4^2}-\frac{1}{4^2}+...+\frac{4^{2009}}{4^{2009}}-\frac{1}{4^{2009}}+\frac{4^{2010}}{4^{2010}}-\frac{1}{4^{2010}}\)
\(C=\left(1+1+...+1\right)-\left(\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}+\frac{1}{4^{2010}}\right)\)(tổng có 2010 số 1)
\(C=2010-\left(\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}+\frac{1}{4^{2010}}\right)\)
Xét tổng \(A=\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}+\frac{1}{4^{2010}}\)
=> \(4A=1+\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}\)
=> \(4A-A=\left(1+\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}\right)-\)\(\left(\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}+\frac{1}{4^{2010}}\right)\)
=> \(3A=1-\frac{1}{4^{2010}}<1\)
=> \(A<\frac{1}{3}\)
=> \(C=2010-A>2010-\frac{1}{3}>2010-1>2009\)
\(C=\frac{4^1-1}{4^1}+\frac{4^2-1}{4^2}+...+\frac{4^{2009}-1}{4^{2009}}+\frac{4^{2010}-1}{4^{2010}}\)
\(C=\frac{4^1}{4^1}-\frac{1}{4^1}+\frac{4^2}{4^2}-\frac{1}{4^2}+...+\frac{4^{2009}}{4^{2009}}-\frac{1}{4^{2009}}+\frac{4^{2010}}{4^{2010}}-\frac{1}{4^{2010}}\)
\(C=\left(1+1+...+1\right)-\left(\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}+\frac{1}{4^{2010}}\right)\)(có 2010 số 1)
\(C=2010-\left(\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}+\frac{1}{4^{2010}}\right)\)
Xét : \(A=\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}+\frac{1}{4^{2010}}\)
\(4A=1+\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}\)
\(4A-A=\left(1+\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}\right)-\left(\frac{1}{4^1}+\frac{1}{4^2}+...+\frac{1}{4^{2009}}+\frac{1}{4^{2010}}\right)\)
\(3A=1-\frac{1}{4^{2010}}< 1\)
\(A< \frac{1}{3}\)
\(C=2010-A>2010-\frac{1}{3}>2010-1>2009\)
mik làm câu A thôi nha
ta có :
1 - 2009/2010 = 1/2010
1 - 2010/2011 = 1/2011
Phần bù nào bé thì phân số đó lớn .
Vì 1/2010 > 1/2011
Nên 2009/2010 > 2010/2011
Ta thấy hiệu giữa mẫu số và tử số của hai phân số bằng nhau ( = 1 )
Để so sánh hai phân số, ta so sánh các hiệu.
\(1-\frac{2009}{2010}\)và \(1-\frac{2010}{2011}\)
Ta có :
\(1-\frac{2009}{2010}=\frac{2010}{2010}-\frac{2009}{2010}=\frac{1}{2010}\)
\(1-\frac{2010}{2011}=\frac{2011}{2011}-\frac{2010}{2011}=\frac{1}{2011}\)
Ta thấy :
\(\frac{1}{2010}>\frac{1}{2011}\)
Hay :
\(1-\frac{2009}{2010}>1-\frac{2010}{2011}\)
Vậy \(\frac{2009}{2010}< \frac{2010}{2011}\)
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
a) Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\) ; \(\frac{1}{3^2}< \frac{1}{2.3}\) ; \(\frac{1}{4^2}< \frac{1}{3.4}\) ; ... ; \(\frac{1}{2010^2}< \frac{1}{2009.2010}\)
=> \(Vt< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}< 1\)
2009/2010=1-1/2010<1-1/2011=2010/2011
vậy 2009/2010<2010/2011
3^400=(3^4)^100=81^100>64^100=4^300
=>1/3^400<1/4^300
Vậy 1/3^400<1/4^300
\(B=\frac{2008+2009+2010}{2009+2010+2011}\)
\(=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
\(< \frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}=A\)
1) \(P=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{11}{5^{12}}\)
\(5P=\frac{1}{5^1}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{11}{5^{11}}\)
\(5P-P=\frac{1}{5^1}+\left(\frac{2}{5^2}-\frac{1}{5^2}\right)+\left(\frac{3}{5^3}-\frac{2}{5^3}\right)+...+\left(\frac{11}{5^{11}}-\frac{10}{5^{11}}\right)-\frac{11}{5^{12}}\)
\(4P=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{11}}-\frac{11}{5^{12}}\)
Đặt \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{11}}\)
\(5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\)
\(5A-A=1+\frac{1}{5}-\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^2}+...+\frac{1}{5^{10}}-\frac{1}{5^{11}}\)
\(4A=1-\frac{1}{5^{11}}\Rightarrow A=\frac{1-\frac{1}{5^{11}}}{4}\)
\(4P=\frac{1-\frac{1}{5^{11}}}{4}-\frac{11}{5^{12}}=\frac{1-\frac{1}{5^{11}}}{16}-\frac{11}{5^{12}\cdot4}< \frac{1}{16}\)