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Ta có : A = (3 + 1) (32 + 1) (34 + 1) ... (364 + 1)
=> 8A = (32 - 1)(32 + 1)(34 + 1)......(364 + 1)
=> 8A = (34 - 1)(34 + 1)......(364 + 1)
=> 8A = (364 - 1)(364 + 1)
=> A = \(\frac{3^{64}-1}{8}\)
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 + 1) (32 + 1) (34 + 1) ... (364 + 1)
2A = (3 - 1)(3 + 1) (32 + 1) (34 + 1) ... (364 + 1)
2A = (32 - 1)(32 + 1) (34 + 1) ... (364 + 1)
= (34 - 1)(34 + 1) ... (364 + 1)
= (38 - 1)(38 + 1)(316+1)(332+1)(364+1)
= (316-1)(316+1)(332+1)(364+1)
= (332-1)(332+1)(364+1)
= (364-1)(364+1)
= (3128-1)
=> A = \(\frac{3^{128}-1}{2}\)
\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
áp dụng hằng đẳng thức \(a^2-b^2\)
ta có 2A=\(3^{128}-1\)=>A=\(\frac{3^{128}-1}{2}\)
B=3.(2^2+1)(2^4+1)...(2^64+1)
=(2^2-1)(2^2+1)(2^4+1)...(2^64+1)
=(2^4-1)(2^4+1)...(2^64+1)
=(2^8-1)...(2^64+1)
.......
=(2^64-1)(2^64+1)
=2^128-1
\(A=\left(3+1\right)\left(3^2+1\right)...\left(3^{64}+1\right)\)
\(\Leftrightarrow\left(3-1\right)A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^{64}+1\right)\)
\(\Leftrightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Leftrightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)
\(\Leftrightarrow2A=3^{128}-1\)
\(\Leftrightarrow A=\frac{3^{128}-1}{2}\)
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
Rút gọn: (3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
A=(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
2A=2(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
2A=(3-1)(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
2A=(3^2-1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
2A=(3^4-1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
2A=(3^8-1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
2A=(3^16-1)(3^16 + 1)(3^32 + 1)
2A=(3^32 - 1)(3^32 + 1)
2A=3^64-1
=>A=(3^64-1) /2
Lời giải :
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\cdot\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\cdot\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\cdot\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\cdot\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\cdot\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\cdot\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(=\frac{1}{2}\cdot\left(3^{64}-1\right)\)
\(=\frac{3^{64}-1}{2}\)
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
(1+3)(1+3^2)(1+3^4)……(1+3^64)+1
=1/2*2*(1+3)(1+3^2)(1+3^4)……(1+3^64)+1
=1/2*(3-1)*(1+3)(1+3^2)(1+3^4)……(1+3^64)+1
=1/2*(3^2-1)(1+3^2)(1+3^4)……(1+3^64)+1
=1/2*(3^4-1)(1+3^4)……(1+3^64)+1
=1/2*(3^8-1)……(1+3^64)+1
..........
=1/2*(3^64-1)(1+3^64)+1
=1/2*(3^128-1)+1
=1/2*3^128-1/2+1
=1/2*3^128+1/2
=1/2*(3^128+1)
\(=\dfrac{3^{128}+1}{2}\)