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đặt A = (2 + 1)(22 + 1)...(2256 + 1).
khi đó (2 - 1)A = (2 -1)(2 + 1)(22 + 1)...(2256 + 1)
suy ra A = 2257 - 1 (dùng hiệu hai bình phương).
nên biểu thức đã cho là A + 1 = 2257.
A = (22 - 1) (22 +1)(24 +1)...(264 +1) + 1 = (24 - 1)(24 +1)...(264 +1) + 1 = (28 -1)...(264 +1) + 1 = 2128 -1 + 1 = 2128
\(A=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\)
\(=2^{64}-1\)
A = 3( 22 + 1 )( 24 + 1 )( 28 + 1 )( 216 + 1 )( 232 + 1 )
= ( 22 - 1 )( 22 + 1 )( 24 + 1 )( 28 + 1 )( 216 + 1 )( 232 + 1 )
= ( 24 - 1 )( 24 + 1 )( 28 + 1 )( 216 + 1 )( 232 + 1 )
= ( 28 - 1 )( 28 + 1 )( 216 + 1 )( 232 + 1 )
= ( 216 - 1 )( 216 + 1 )( 232 + 1 )
= ( 232 - 1 )( 232 + 1 )
= 264 - 1
3(22 + 1)(24 + 1)(28 + 1)(216 + 1)
=(4 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
=(22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
=(24 - 1)(24 + 1)(28 + 1)(216 + 1)
=(28 - 1)(28 + 1)(216 + 1)
=(216 - 1)(216 + 1)
=(232 - 1)
Đặt A=3(22 +1)(24+1)(28+1)(216+1)
=(4-1)(22+1)(24+1)(28+1)(216+1)
=[(22-1)(22+1)](24+1)(28+1)(216+1)
=(24-1)(24+1)(28+1)(216+1)
=(28-1)(28+1)(216+1)
=(216-1)(216+1)
=232-1
3(22 +1)(24+1)(28+1)(216+1) = (22 -1)(22 +1)(24+1)(28+1)(216+1) = (24-1)(24+1)(28+1)(216+1) = (28-1)(28+1)(216+1)
= (216-1)(216+1) = 232-1
( bài này áp dụng hằng đẳng thức \(a^2-b^2=\left(a+b\right)\left(a-b\right)\)
Ta có
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\)
\(=2^{64}-1\)
3.(22+1)(24+1)(28+1)(216+1)(232+1)
=(22-1)(22+1)(24+1)(28+1)(216+1)(232+1)
=(24-1)(24+1)(28+1)(216+1)(232+1)
=(28-1)(28+1)(216+1)(232+1)
=(216-1)(216+1)(232+1)
=(232-1)(232+1)
=264-1
3(22+1)(24+1)(28+1)(216+1)(232+1)(264+1)
=(22-1)(22+1)(24+1)(28+1)(216+1)(232+1)(264+1)
=(24-1)(24+1)(28+1)(216+1)(232+1)(264+1)
=(28-1)(28+1)(216+1)(232+1)(264+1)
=(216-1)(216+1)(232+1)(264+1)
=(232-1)(232+1)(264+1)
=(264-1)(264+1)
=(2128-1)
Nếu thấy đúng thì thích cho mình nha
A=(2-1)(2+1)*...*(2^256+1)+1
=(2^2-1)(2^2+1)*...*(2^256+1)+1
=(2^4-1)(2^4+1)*...*(2^256+1)+1
=(2^8-1)(2^8+1)(2^16+1)(2^32+1)*....*(2^256+1)+1
=(2^16-1)(2^16+1)*....*(2^256+1)+1
=(2^32-1)(2^32+1)*...*(2^256+1)+1
=(2^64-1)(2^64+1)(2^128+1)(2^256+1)+1
=(2^128-1)(2^128+1)(2^256+1)+1
=(2^256-1)(2^256+1)+1
=2^512