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Ta có:
P=12(52+1)(54+1)(58+1)(516+1)
P=(52-1)(52+1)(54+1)(58+1)(516+1):2
P=(54-1)(54+1)(58+1)(516+1):2
P=(58-1)(58+1)(516+1):2
P=(516-1)(516+1):2
P=(532-1):2
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)=\frac{5^{32}-1}{2}\)
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
a/ \(\Rightarrow A=\frac{x^2+x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x+1}{x^2+x+1}-\frac{1}{x-1}\)
\(=\frac{x^2+x+\left(x+1\right)\left(x-1\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{x^2+x+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{x^2-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
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)
( 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
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
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
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.