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A=\(2^0+2^1+2^2+2^3+...+\)\(2^{2010}\)
2A=\(2^1+2^2+2^3+...+2^{2011}\)
2A-A=\(2^1+2^2+2^3+...+2^{2011}\)-\(2^0-2^1-2^2-...-2^{2010}\)
A=\(2^{2011}-1\)
vậy A=\(2^{2011}-1\)
\(A=2^0+2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=1+2^1+2^2+2^3+2^4+...+2^{2010}\)
=> \(2A=2+2^2+2^3+2^4+2^5+...+2^{2011}\)
=> \(A=2^{2011}-1\)
Study well ! >_<

a, \(A=2^0+2^1+2^2+...+2^{2010}\)
\(=>2A=2^1+2^2+2^3+...+2^{2011}\)
\(=>2A-A=\left(2^1+2^2+2^3+...+2^{2011}\right)-\left(2^0+2^1+2^2+...+2^{2010}\right)\)
\(=>2A=2^{2011}-2^0=2^{2011}-1\)
Vì \(2^{2011}-1=2^{2011}-1\)
\(=>A=B\)
a) Ta có : A=1+2+22+...+22010
2A=2+22+23+...+22011
\(\Rightarrow\) 2A-A=(2+22+23+...+22011)-(1+2+22+...+22010)
\(\Rightarrow\) A=22011-1
Mà B=22011-1
\(\Rightarrow\)A=B
Vậy A=B.
b) Ta có : A=2009.2011
B=20102=2010.2010
\(\Rightarrow\)A=2009.2010+2009
B=2009.2010+2010
Vì 2009<2010 nên 2009.2010+2009<2009.2010+2010
hay A<B
Vậy A<B.


z lm lại, nhưng nó cứ sao ý!
\(\left(2^2+2^1+2^2+2^3\right).2^0.2^1.2^2.2^3\)
\(=18.64\)
\(=1152\)
Sửa đề:
đặt A = \(\left(2^0+2^1+2^2+2^3\right).2^0.2^1.2^2.2^3\)
\(A=\left(1+2+2^2+2^3\right).2^6\)
\(A=2^6+2^7+2^8+2^9\)
\(\Rightarrow2A=2^7+2^8+2^9+2^{10}\)
\(\Rightarrow2A-A=2^{10}-2^6\)
\(A=2^{10}-2^6\)
\(A=960\)
xem đề mk ghi như z cs đúng ko?

a, 5x - 1 = 13
=> 5x = 14
=> x = 14/5
b,(x - 2) = 0
=> x - 2 = 0
=> x = 2
c, 5(x - 7) + 8 = 0
=>5(x - 7) = -8
=>x -7 = -8/5 = -1,6
=>x = 5,4
d, (x - 19).4 = 36
=>x - 19 = 9
=>x = 28
e, 3(x - 7) - 2 = 4
=> 3(x - 7) = 6
=> x - 7 = 2
=> x = 9

\(2^2.2^5-\left(2010^0+19\right):2^2\)
\(=2^7.20:2^2\)
\(=2560:2^2\)
\(=640\)
2^2.2^5-(2010^0+19):2^2
=2^7 - (1+19):4
=128- 20:4
=128 - 5
=123