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\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\left(3^{64}-1\right)\)
A=\(\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^{32}-1\right)\left(3^{32}+1\right)=3^{64}-1\)
đặt A=(3+1)(32+1)(34+1)(38+1)(316+1)(332+1)
=>2A=2.(3+1)(32+1)(34+1)(38+1)(316+1)(332+1)
=>2A=(3-1)(3+1)(32+1)(34+1)(38+1)(316+1)(332+1)
=(32-1)(32+1)(34+1)(38+1)(316+1)(332+1)
=(34-1)(34+1)(38+1)(316+1)(332+1)
=(38-1)(38+1)(316+1)(332+1)
=(316-1)(316+1)(332+1)
=(332-1)(332+1)
=364-1
=>2A=\(\frac{3^{64}-1}{2}\)
bạn ơi: A=\(\frac{3^{64}-1}{2}\) chứ ko phải 2A=\(\frac{3^{64}-1}{2}\)
Ta có A = (32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)(364 + 1)
=> 8A = 8(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)(364 + 1)
=> 8A = (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)(364 + 1)
=> 8A = (34 - 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)(364 + 1)
=> 8A = (38 - 1)(38 + 1)(316 + 1)(332 + 1)(364 + 1)
=> 8A = (316 - 1)(316 + 1)(332 + 1)(364 + 1)
=> 8A = (332 - 1)(332 + 1)(364 + 1)
=> 8A = (364 - 1)(364 + 1)
=> 8A = 3128 - 1 (1)
Đặt B = 3126
=> 8B = 3126 . 8 = 3126.(32 - 1) = 3128 - 3126 (2)
Từ (1)(2) => 8A > 8B
=> A > B
\(8.\left(3^2+1\right).\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)-3^{32}\)
\(=\left(3^2-1\right).\left(3^2+1\right).\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)-3^{32}\)
\(=\left(3^4-1\right).\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)-3^{32}\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)-3^{32}\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)-3^{32}=3^{32}-1-3^{32}=-1\)
A = (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)
=>2A=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
=>A<B
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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=\(\frac{3^{32}-1}{2}<3^{32-1}\)
Vậy A<B
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
Đặt biểu thức đó là A
(3-1) A= (3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1) (3^32+1)
2 A= (3^2-1)(3^2+1)(3^4+1)..............................................
2A = (3^4-1)(3^4+1)(3^8+1) ............................
2A= (3^8-1)(3^8+1)(3^16+1) .............
2A = (3^16-10(3^16+1)(3^32+1)
2A = (3^32-1)(3^32+1)
2A= 3^64-1
A= (3^64-1) / 2
\(A=\left(3+1\right)\left(3^2+1\right)....\left(3^{16}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)....\left(3^{16}+1\right)\)\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)....\left(3^{16}+1\right)\)\(2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(2A=\left(3^{32}-1\right)=B\Rightarrow B>A\)
Theo bài ra ta có:
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ \Rightarrow2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ \Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ \Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ \Rightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ \Rightarrow2A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ \Rightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\\ \Rightarrow2A=3^{32}-1\\ \Rightarrow A=\dfrac{3^{32}-1}{2}\) ta thấy : \(\dfrac{3^{32}-1}{2}< 3^{32}-1\\ \)
=> A < B
vậy A < B