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D
0
23 tháng 10 2017
Do : b + 1 = a --> a - b = 1
Ta có : ( a + b)( a2 + b2)( a4 + b4)( a8 + b8)( a16 + b16)
= 1.( a + b)( a2 + b2)( a4 + b4)( a8 + b8)( a16 + b16)
= ( a - b)( a + b)( a2 + b2)( a4 + b4)( a8 + b8)( a16 + b16)
= ( a2 - b2)( a2 + b2)( a4 + b4)( a8 + b8)( a16 + b16)
= ( a4 - b4)( a4 + b4)( a8 + b8)( a16 + b16)
= ( a8 - b8)( a8 + b8)( a16 + b16)
= ( a16 - b16)( a16 + b16)
= a32 - b32 ( đpcm)
NL
1
DN
26 tháng 4 2016
(a + b)(a2 + b2)(a4 + b4)(a8 + b8)(a16 + b16)
=1.(a + b)(a2 + b2)(a4 + b4)(a8 + b8)(a16 + b16)
= (a – b) (a + b)(a2 + b2)(a4 + b4)(a8 + b8)(a16 + b16)
= (a2 – b2) (a2 + b2)(a4 + b4)(a8 + b8)(a16 + b16)
= (a4 – b4)(a4 + b4)(a8 + b8)(a16 + b16)
= (a8 – b8)(a8 + b8)(a16 + b16)
= (a16– b16)(a16 + b16)
= a32 – b32
HH
2
2 tháng 10 2015
A=(3+1)(32+1)(34+1)(38+1)(316+1)
=>2A=(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=B
=>B=1.A
=>k=1
Vậy k=1
2 tháng 10 2015
Ta có :A=(3+1)(32+1)(34+1)(38+1)(316+1)
2A=2.(3+1)(32+1)(34+1)(38+1)(316+1)
2A=(3-1)(3+1)(32+1)(34+1)(38+1)(316+1)
2A=(32-1)(32+1)(34+1)(38+1)(316+1)
2A=(34-1)(34+1)(38+1)(316+1)
2A=(38-1)(38+1)(316+1)
2A=(316-1)(316+1)
2A=332-1
Lại có :B=332-1 =2A =>k=2
HN
1
9 tháng 9 2018
B = ( 3 + 1 ).( 32 + 1 ).(34+1).(38+1).(316+1)
=> 2B = 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 = 332-1:2<332-1
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
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
29 tháng 7 2017
a, \(A=1999.2001=\left(2000-1\right)\left(2000+1\right)=2000^2-1< 2000^2=B\)
Vậy A<B
b, \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1< 2^{32}=B\)
Vậy A<B
29 tháng 7 2017
a, \(A=1999.2001=\left(2000-1\right)\left(2000+1\right)\)
\(=2000^2-1< 2000^2\)
\(\Rightarrow A< B\)
b, \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1< 2^{32}\)
\(\Rightarrow A< B\)
Phân tích vế trái ta có:
\(a^{32}-b^{32}=\left(a^{16}\right)^2-\left(b^{16}\right)^2\)
\(=\left(a^{16}+b^{16}\right)\left(a^{16}-b^{16}\right)\)
\(=\left(a^{16}+b^{16}\right)\left(\left(a^8\right)^2-\left(b^8\right)^2\right)\)
\(=\left(a^{16}+b^{16}\right)\left(a^8+b^8\right)\left(a^8-b^8\right)\)
Tượng tự ta có :
\(=\left(a^{16}+b^{16}\right)\left(a^8+b^8\right).....\left(a^4+b^4\right)\left(a^4-b^4\right)\)
\(=\left(a^{16}+b^{16}\right).....\left(a^4+b^4\right)\left(a^2+b^2\right)\left(a^2-b^2\right)\)
\(=\left(a^{16}+b^{16}\right).......\left(a^2+b^2\right)\left(a+b\right)\left(a-b\right)\)
Do a-b=1 nên
\(=>a^{32}-b^{32}=\left(a^{16}+b^{16}\right)....\left(a^4+b^4\right)\left(a^2+b^2\right)\left(a+b\right)\)
CHÚC BẠN HK TỐT............