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PN
3
MT
9 tháng 7 2015
B=(2+1)(22+1)(24+1)(28+1)(216+1)−232
=1.(2+1)(22+1)(24+1)(28+1)(216+1)−232
=(2-1)(2+1)(22+1)(24+1)(28+1)(216+1)−232
=(22-1)(22+1)(24+1)(28+1)(216+1)−232
=(24-1)(24+1)(28+1)(216+1)−232
=(28-1)(28+1)(216+1)−232
=(216-1)(216+1)−232
=232-1-232
=-1
9 tháng 7 2015
A = ( 2 +1 )( 2^2 + 1 )...(2^16+1) - 2^32
A = ( 2 - 1) ( 2 + 1 )(2^2 + 1) .... (2^16 + 1) - 2^32
A = (2^2 - 1) (2^2 + 1) ...(2^16 + 1) - 2^32
A =( 2^ 4 - 1)( 2^4 + 1 )( 2^8 + 1) (2^16+1) -2^32
A = ( 2^8 - 1)( 2^ 8 + 1) ( 2^ 16 + 1)- 2^32
A = ( 2^16 - 1 )( 2^16 + 1) - 2^32
A = 2^32 - 1 - 2^32
A = - 1
NT
1
DN
24 tháng 10 2018
\(A=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Rightarrow A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Rightarrow A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Rightarrow A=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Rightarrow A=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Rightarrow A=\left(2^{32}-1\right)\left(2^{32}+1\right)\)
\(\Rightarrow A=2^{64}-1\)
Vậy \(A=2^{64}-1\)
NT
1
5 tháng 11 2018
\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(A=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(A=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(A=\left(2^{32}-1\right)\left(2^{32}+1\right)\)
\(A=2^{64}-1\)
6 tháng 7 2016
A= \(\frac{3\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{\left(2^2-1\right)}=2^{32-1}\)
mà B= \(2^{32}\)
=> A<B
CE
11 tháng 9 2016
Nhân với 2-1 áp dụng bất đẳng thức a^2-b^2=(a-b)(a+b)
=> 2^64-1
11 tháng 9 2016
(2+1)(22+1)(24+1)(28+1)(216+1)(232+1)
=[3(22+1)(24+1)](28+1)(216+1)(232+1)
=[(22-1)(22+1)](24+1)(28+1)(216+1)(232+1)
=[(24-1)(24+1)](28+1)(216+1)(232+1)
=[(28-1)(28+1)](216+1)(232+1)
=[(216-1)(216+1)](232+1)
=(232-1)(232+1)
21 tháng 9 2016
(2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1) - 232
= (2 - 1)(2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1) - 232
= (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1) - 232
= (24 - 1)(24 + 1)(28 + 1)(216 + 1) - 232
= (28 - 1)(28 + 1)(216 + 1) - 232
= (216 - 1)(216 + 1) - 232
= (232 - 1) - 232
= 232 - 1 - 232
= -1
DP
1
TT
17 tháng 9 2018
A = 12 – 22 + 32 – 42 + … – 20042 + 20052
A = 1 + (32 – 22) + (52 – 42)+ …+ ( 20052 – 20042)
A = 1 + (3 + 2)(3 – 2) + (5 + 4 )(5 – 4) + … + (2005 + 2004)(2005 – 2004)
A = 1 + 2 + 3 + 4 + 5 + … + 2004 + 2005
A = ( 1 + 2002 ). 2005 : 2 = 2011015
b/ B = (2 + 1)(22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = (22 - 1) (22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = ( 24 – 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = …
B =(232 - 1)(232 + 1) – 264
B = 264 – 1 – 264
B = - 1
TT
17 tháng 9 2018
xin lỗi nha chỗ câu a mình lộn
chỗ (1+2002)x2005:2=2011015 là sai nha
(1+2005)x2005:2= 2011015 là đúng nha
a) Ta có: \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)