DH
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K
Khách
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ZD
1
30 tháng 6 2016
Áp dụng đẳng thức : a^2 - 1 = (a + 1)(a - 1)
=> a + 1 = (a^2 - 1)/(a + 1)
Ta có: 3 + 1 = (3^2 - 1)/(3 - 1)
3^2 + 1 = (3^4 - 1)/(3^2 - 1)
3^4 + 1 = (3^8 - 1)/(3^4 - 1)
3^8 + 1 = (3^16 - 1)/(3^8 - 1)
3^16 + 1 = (3^32 - 1)/(3^16 - 1)
3^32 + 1 = (3^64 - 1)/(3^32 - 1)
(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
=(3^2 - 1)/(3 - 1).(3^4 - 1)/(3^2 - 1).(3^8 - 1)/(3^4 - 1).(3^32 - 1)/(3^16 - 1).(3^64 - 1)/(3^32 - 1)
=(3^64 - 1)/(3 - 1)
=(3^64 - 1)/2
TN
1
PN
0
NH
2
HT
3 tháng 9 2017
a) A = 2016.2018 = ( 2017 - 1 ).2018 = 2017.2018 - 2018 ( 1 )
B = 20172 = 2017.2017 = 2017.( 2018 - 1) = 2017.2018 - 2017 ( 2 )
Từ (1) và (2), ta thấy: - 2018 < - 2017 => 2017.2018 - 2018 < 2017.2018 - 2017 <=> 2016.2018 < 20172
Vậy A < B
~ Phần b khi nào nghĩ ra tớ sẽ làm ngay ạ :) Còn phần này chắc chắn đúng cậu nhé ~
TN
3 tháng 9 2017
b)\(x=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2x=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2x=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2x=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2x=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2x=\left(3^{16}-1\right)\left(3^{16}+1\right)\Rightarrow x=\frac{3^{32}-1}{2}\)
Thấy \(x=\frac{3^{32-1}}{2}< 3^{32}-1=y\)
ML
3
16 tháng 6 2016
B=(3+1).(3^2+1).(3^4+1).(3^8+1).(3^16 +1).2
=>B=2.(3+1)(32+1)(34+1)(38+1).(316+1)
=(3-1)(3+1)(32+1)(34+1)(38+1).(316+1)
=(32-1)(32+1)(34+1)(38+1).(316+1)
=(34-1)(34+1)(38+1).(316+1)
=(38-1)(38+1).(316+1)
=(316-1)(316+1)
=332-1=A
Vậy A=B
16 tháng 6 2016
\(B=2.\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3-1\right).\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3^2-1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3^4-1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3^8-1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right).\left(3^{16}+1\right)\)
\(=3^{32}-1\)
Vậy A = B = 332 - 1
LN
1
KB
14 tháng 9 2018
Ta có : \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3^{64}-1\right)\)
\(=\dfrac{3^{64}-1}{2}\)
:D
LQ
3
23 tháng 7 2018
\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2B=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2B=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2B=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(B=\dfrac{3^{32}-1}{2}< A=3^{32}-1\)
23 tháng 7 2018
\(B=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =>2B=2.\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^{16}-1\right)\left(3^{16}+1\right)=3^{32}-1\\ =>A=\dfrac{3^{32}-1}{2}< B\)
TD
3
MA
7 tháng 10 2016
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\left(3^{64}-1\right)\)
KK
0
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(2A=3^{64}-1\Rightarrow A=\dfrac{3^{64}-1}{2}\)