so sanh 2^1+2^2+2^3+...+2^2010 va 2^2010+1
So sanh :2009^2009+1/2009^2010+1 va 2009^2010-2/2009^2011-2
So sanh : .A=2^0+2^1+2^2+2^3+...+2^2010 va` B=2^2011-1.
Có A=20+21+22+23+24+.....+22010
Nên 2A = 2 (20+21+22+23+24+.....+22010 )
= 21+22+23+24+.....+22011 + 22011
=>A = 2A - A = 22011 - 20
= 22011 - 1
= B
Vậy A = B
H = 1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + ... + 2^2009
k = 2^2010 - 1
So sanh k va H
\(H=1+2+2^2+2^3+2^4+....+2^{2009}\)
\(\Rightarrow2H-H=\left(2+2^2+2^3....+2^{2010}\right)-\left(1+2+2^2+...+2^{2009}\right)\)
\(\Rightarrow H=2^{2010}-1=K\)
2.H = 2 + 22 + 23 + ... + 22010
2.H - H = (2 + 22 + 23 + ... + 22010) - (1 + 2 + 22 + ... + 22009)
H = 22010 - 1 = k
Bai 1.Tim x, y biet :
2x(3y-2)+(3y-2) = -55
Bai 2 .a) So sanh : -22/45 va -51/103
b) So sanh A = 2009^2009 +1 / 2009^2010 va B = 2009^2010-2/2009^2011-2
Bai 3 :
a)Tim so tu nhien co 3 chu so , biet rang khi chia so do cho cac so 25, 28,35thi duoc cac so du lan luot la 5,8,15
b)Tim x: (x+1)+(x+2)+(x+3)+...+(x+100)=205550
So sanh : U = \(\frac{2009^{2005}+1}{2009^{2010}+1}\)va V = \(\frac{2009^{2010}+2}{2009^{2011}+2}\)
cho A= 3+2^2+2^3+...+2^2010+2^2011,B=2^2012.so sanh A va B
cho A= 3+2^2+2^3+...+2^2010+2^2011,B=2^2012.so sanh A va B
2A = 6+2^3+2^4+.....+2^2012
A = 2A - A = (6+2^3+2^4+.....+2^2012)-(3+2^2+2^3+......+2^2011)
= 6+2^2012 - 3 - 2^2
= 2^2012 - 1
=> A < B
Tk mk nha
ta có :
\(A=3+2^2+2^3+.....+2^{2011}.\)
\(\Rightarrow2A=6+2^3+2^4+....+2^{2012}\)
\(\Rightarrow A=\left(6+2^3+2^4+...+2^{2012}\right)-\left(3+2^2+2^3+....+2^{2011}\right)\)
\(\Rightarrow A=-1+2^{2012}\)
vì -1+2^2012<2^2012 nên A <B
so sanh:2^2010+1/2^2007+1 va 2^2012=1/2^2009+1
so sanh
\(\frac{2^{2010}+1}{2^{2007}+1}va\frac{2^{2012}+1}{2^{2009}+1}\)