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Ta có: (2+1)(22+1)(24+1)(28+1)(216+1)(232+1)
= (2-1)(2+1)(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
A = 12 – 22 + 32 – 42 + … – 20042 + 20052
A = 1 + (32 – 22) + (52 – 42)+ …+ ( 20052 – 20042)
A = 1 + (3 + 2)(3 – 2) + (5 + 4 )(5 – 4) + … + (2005 + 2004)(2005 – 2004)
A = 1 + 2 + 3 + 4 + 5 + … + 2004 + 2005
A = ( 1 + 2002 ). 2005 : 2 = 2011015
b/ B = (2 + 1)(22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = (22 - 1) (22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = ( 24 – 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = …
B =(232 - 1)(232 + 1) – 264
B = 264 – 1 – 264
B = - 1
xin lỗi nha chỗ câu a mình lộn
chỗ (1+2002)x2005:2=2011015 là sai nha
(1+2005)x2005:2= 2011015 là đúng nha
\(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
b/ \(100^2+\left(100+3\right)^2+\left(100+5\right)^2+\left(100-6\right)^2\)
\(=100^2+100^2+100^2+100^2+4.100+9+25+36\)
\(=100^2+2.100+1+100^2-4.100+4+100^2-8.100+16+100^2+14.100+49\)
\(=\left(100+1\right)^2+\left(100-2\right)^2+\left(100-4\right)^2+\left(100+7\right)^2\)
\(A=-1^2+2^2-3^2+4^2-...-99^2+100^2\)
\(=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(100^2-99^2\right)\)
\(=\left(2+1\right)\left(2-1\right)+\left(4+3\right)\left(4-3\right)+...+\left(100+99\right)\left(100-99\right)\)
\(=1+2+3+4+...+99+100\)
\(=\frac{\left(1+100\right)\cdot100}{2}=5050\)
\(C=\left(2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}=\left(2^{64}-1\right)-2^{42}=-1\)
Mk chỉ bt làm câu C thôi tại vì mk chỉ học lớp 7
C=(2+1)(24+1)(28+1)(216+1)(232+1)-264
C=(24-1)(24+1)(28+1)(216+1)(232+1)-264
C=(28-1)(28+1)(216+1)(232+1)-264
C=(216-1)(216+1)(232+1)-264
C=(232-1)(232+1)-264
C=264-1-264
C=-1
Nhân với 2-1 áp dụng bất đẳng thức a^2-b^2=(a-b)(a+b)
=> 2^64-1
(2+1)(22+1)(24+1)(28+1)(216+1)(232+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)
(2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1) - 232
= (2 - 1)(2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1) - 232
= (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1) - 232
= (24 - 1)(24 + 1)(28 + 1)(216 + 1) - 232
= (28 - 1)(28 + 1)(216 + 1) - 232
= (216 - 1)(216 + 1) - 232
= (232 - 1) - 232
= 232 - 1 - 232
= -1
Ta có : ( 2+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 (đpcm)