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LD
3
16 tháng 9 2019
\(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(A=\left(2^8-1\right)\left(2^8+1\right)\)
\(A=2^{16}-1< 2^{16}\)
16 tháng 9 2019
Mình làm theo cách tính nhé !
\(A=\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right)\)
\(A=3.\left(4+1\right).\left(16+1\right).\left(256+1\right)\)
\(A=3.5.17.257\)
\(\Rightarrow A=65535\)
\(B=2^{16}=65536\)
Từ đó \(\Rightarrow A< B\)
QM
0
VT
2
19 tháng 7 2015
A = (2 - 1)(2 + 1)(2^2 + 1 )(2^4 + 1 ) (2^8 + 1)(2^16 + 1) ( nhân vói 2 - 1 = 1 Gía không thay dổi)
A = ( 2 ^2 - 1 )(2^2 + 1 )(2^4 + 1 )(2^8 + 1 )(2^16 + 1 )
A = ( 2^4 - 1 )(2^4 + 1)(2^8 + 1)(2^16 + 1)
A = (2^8 - 1)(2^8 + 1)(2^16 + 1)
A = (2^16 - 1)(2^16 + 1 )
A = 2^32 - 1 <2^32 = B
VẬy A < B
PV
1
27 tháng 7 2016
a)
A = 1999.2001 = (2000-1)(2000+1)=20002-1
vì 20002 -1 < 20002 nên A<B
10 tháng 11 2017
\(\text{a) }\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\dfrac{3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ \\ =\dfrac{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{\left(2^{16}-1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{2^{32}-1}{3}\\ \)
\(\text{b) }24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\\ =\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\\ =\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right) \\ =\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\\ =\left(5^{16}-1\right)\left(5^{16}+1\right)\\ =5^{32}-1\\ \)
\(\text{c) }48\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^4-1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^8-1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^{16}-1\right)\left(7^{16}+1\right)\\ =7^{32}-1\)
4 tháng 7 2017
Ta có:
a) A = 2018 x 2020 = (2019 - 1) x (2019 + 1)
Áp dụng hằng đẳng thức thứ ba ta có:
A = 208 x 2020 = \(2019^2-1^2=2019^2-1\)
Vì \(2019^2-1< 2019^2\)
\(\Rightarrow\)A < B
b) A = \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1^2\right)\left(2^2+1^2\right)\left(2^4+1^2\right)\left(2^8+1^2\right)\left(2^{16}+1^2\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
Vì \(2^{32}-1< 2^{32}\)
\(\Rightarrow\)A < B
4 tháng 7 2017
a) Áp dụng hàng đăng thức (a - b) (a + b) = a2 - b2
Ta có : A = 2018.2020 = (2019 - 1) (2019 + 1) = 20192 - 1
Mà B = 20192
Nên A < B
ND
2
10 tháng 8 2019
B=\(2^{16}-1\)
\(A=2+1.2^2+1.2^4+1.2^8+1\)\(=\left(2.2^2.2^4.2^8\right)+\left(1+1+1+1\right)\)\(=2^{15}+4\)
mà \(2^{16}>2^{15}\)=> A>B
NH
3
A=65536 > B=65535 do:
A=\(2^{16}\)
=65536
B=\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
=3x5x17x257
=65535