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Lời giải:
Xét tử số:
$X=1+2+2^2+2^3+...+2^{2008}$
$2X=2+2^2+2^3+2^4+....+2^{2009}$
$\Rightarrow 2X-X=(2+2^2+2^3+2^4+....+2^{2009})-(1+2+2^2+...+2^{2008})$
$\Rightarrow X=2^{2009}-1$
$\Rightarrow S=\frac{X}{1-2^{2009}}=\frac{2^{2009}-1}{-(2^{2009}-1)}=-1$
\(B=2^{2010}-2^{2009}+2^{2008}-...+2^2-2\)
\(2B=2^{2011}-2^{1010}+2^{2009}-...+2^3-2^2\)
\(3A=2^{2011}-2\)
\(A=\frac{2\left(2^{2010}-1\right)}{3}\)
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=\(\frac{1+2+2^2+...+2^{2008}}{1-2^{2009}}\)=\(\frac{2+2^2+2^3...+2^{2009}-1-2-2^2-...-2^{2008}}{\left(1-2^{2009}\right)}\)=\(\frac{2^{2009}-1}{1-2^{2009}}\)=-1
Vậy: B=-1
a, 33.( 17- 5) - 17.( 33-5)
= 33.17 - 33.5 - 17.33 + 17.5
= ( 33.17 - 17.33) - ( 33.5 - 17.5)
= 0 - 5.( 33- 17)
= - 5. 16
= - 80
b, 12 + 3.{ 90 : [ 39 - ( 23 - 5)2]
= 12 + 3. { 90 : [ 39 - ( 8-5)2 ]}
= 12 + 3 . { 90 : [ 39 - 32 ]}
= 12 + 3.{ 90 : (39 -9)}
= 12 + 3. { 90 : 30}
= 12 + 3 . 3
= 12 + 9
= 21
c, 307 - [ (180 .40 - 160 ) : 22 + 9] : 2
= 307 - [ ( 180 - 160) : 4 + 9]:2
= 307 - [ 20:4 +9 ] :2
= 307 - [ 5 + 9] : 2
= 307 - 14 : 2
= 307 - 7
= 300
nhóm như sau:
(1+3+5+....+2009) --- ( 2+4+6+.....+2010)
= {{ ((2009 -1)/2 +1) x (2009 +1) } / 2 }} --- {{ (( 2010 - 2) /2+1) x (2010+2)) / 2 }}
= 1010025 --- 1011030
= -1005
giá trị rút gọn là sao pạn?mình ko pit mình chỉ pit kết quả:
C=1-2+3-4+....+2007-2008+2009-2010
=(1-2)+(3-4)+...+(2007-2008)+(2009-2010)
=-1+-1+....+-1+-1
=-1.(-2010-1+1):2
=1.(-2010):2
=1.(-1005)
=-1005
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