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Ta có: \(A=\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=3\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4-1\right)\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^8-1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^{16}-1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^{32}-1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=4^{64}-1\)
mà \(B=4^{64}-1\)
Vậy \(B=3A\)
Ta có (42 - 1)(42 + 1) = 44 - 1
Ta có 15A = (42 - 1)(42 + 1)(44 + 1)(48 + 1)(416 + 1)(432 + 1) - 464 = 464 - 1 - 464 = -1
=> A = \(\frac{-1}{15}\)
Sửa B=432-1
Ta có: \(3A=\left(4-1\right)\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)
\(=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)
\(=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)=\left(4^8-1\right)\left(4^8+1\right)\left(4^{16}+1\right)\)
\(=\left(4^{16}-1\right)\left(4^{16}+1\right)=4^{32}-1=B\) (đpcm)
Bài 2:
Ta có: \(\frac{x+1}{x}=10\) hay \(\frac{x^1+1^1}{x^1}=10^1\)
Nên suy ra : \(\frac{x^5+1}{x^5}=10^5\)
= 100000 ( do 15 cũng sẽ =1 nên không viết mũ 5 cũng chả sao)
A = (2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) - 264
A = (2 - 1)(2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) - 264
A = (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) - 264
A = (24 - 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) - 264
A = (28 - 1)(28 + 1)(216 + 1)(232 + 1) - 264
A = (216 - 1)(216 + 1)(232 + 1) - 264
A = (232 - 1)(232 + 1) - 264
A = 264 - 1 - 264
A = -1
Ta có:\(A=\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4-1\right)\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^8-1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^{16}-1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=\left(4^{32}-1\right)\left(4^{32}+1\right)\)
\(\Rightarrow3A=4^{64}-1\)
\(\Rightarrow3A=B\)
oa