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B=\(\frac{1+2+2^2+...+2^{2008}}{1-2^{2009}}\)=\(\frac{2+2^2+2^3...+2^{2009}-1-2-2^2-...-2^{2008}}{\left(1-2^{2009}\right)}\)=\(\frac{2^{2009}-1}{1-2^{2009}}\)=-1
Vậy: B=-1
Lời giải:
Xét tử số:
$X=1+2+2^2+2^3+...+2^{2008}$
$2X=2+2^2+2^3+2^4+....+2^{2009}$
$\Rightarrow 2X-X=(2+2^2+2^3+2^4+....+2^{2009})-(1+2+2^2+...+2^{2008})$
$\Rightarrow X=2^{2009}-1$
$\Rightarrow S=\frac{X}{1-2^{2009}}=\frac{2^{2009}-1}{-(2^{2009}-1)}=-1$
a, 1004 .2009 + 1005 = (1005-1) .2009 +1005
= 1005 .2009 -2009 +1005
= 1005 .2009 -1004
Vậy ( 1004 .2009 +1005) / (1005 .2009 -1004) =1
b, 1004 .2010 +1 = 1004 .2009 +1004 +1
= (1006 -2) .2009 +1005
= 1006 .2009 -2 .2009 +1005
= 1006 .2009 -4008 +1005
= 1006 .2009 -3013
Vậy (1004 .2010 +1) / (1006 .2009 -3013) = 1
c, 2007 .2009 -2 = 2007.(2008+1) -2
= 2007.2008 +2007 -2
= 2007.2008 +2005
= (2008-1) .2008 +2005
= 2008 .2008 -2008 +2005
= 2008 .2008 -3
Vậy (2008 .2008 -3) / (2007 .2009 -2) =1
\(B=2^{2010}-2^{2009}+2^{2008}-...+2^2-2\)
\(2B=2^{2011}-2^{1010}+2^{2009}-...+2^3-2^2\)
\(3A=2^{2011}-2\)
\(A=\frac{2\left(2^{2010}-1\right)}{3}\)
dễ ợt
s=2010(1+20100+2010^3(1+2010)+............+2010^2009(1+2010)
s=2010.2011+2010^3.2011+.........+2010^2009.2011
s=2011(2010+2010^3+.......+2010^2009) chia hết cho 2011
\(S=\left(2010+2010^2\right)+\left(2010^3+2010^4\right)+...+\left(2010^{2009}+2010^{2010}\right)\)
\(S=2010\left(2010+1\right)+2010^3\left(2010+1\right)+...+2010^{2009}\left(2010+1\right)\)
\(S=2011.\left(2010+2010^3+2010^5+...+2010^{2009}\right)\) chia hết cho 2011
a.|x-1|=3
=>x-1=3 hoặc x-1=-3
=>x=4 hoăc x=-2
vậy...
b.2x+17=15
=>2x=15-17
=>2x=-2
=>x=-1
vậy...
c.tự giải
d.4x-15=-75-x
=>4x+x=-75+15
=>5x=-60
=>x=-12
vậy...
2.2+(-3)+4+...+(-2011)+2012
=(-3+2)+...+(-2011+2010)+2012 (có 2010 cặp)
=-1.2010+2012
=-2010+2012=2
a)2012/2013<2013/2014
b)2003x........< 2007x......
cho mình nha