Ta có 2003.2005=2003.(2004+1)=2003.2004+2003
2004^2=2004.2004=2004.(2003+1)=2003.2004+2004
Vì 2003<2004 nên 2003.2004+2003<2003.2004+2004
Vậy 2003.2005<2004^2
Ta có A=2003.2005=2003.(2004+1)=2003.2004+2003A=2003.2005=2003.(2004+1)=2003.2004+2003
B=20042=2004.2004=2004.(2003+1)=2003.2004+2004B=20042=2004.2004=2004.(2003+1)=2003.2004+2004
Vì 2003<2004 nên 2003.2004+2003<2003.2004+2004
Vậy A<B
tick nha để mk làm câu b
\(Ta có: 8 ( 7 8 + 1 ) ( 7 4 + 1 ) ( 7 2 + 1 ) = 1/ 6 .48 ( 7 2 + 1 ) ( 7 4 + 1 ) ( 7 8 + 1 ) = 1/ 6 ( 7 2 − 1 ) ( 7 2 + 1 ) ( 7 4 + 1 ) ( 7 8 + 1 ) = 1/ 6 ( 7 4 − 1 ) ( 7 4 + 1 ) ( 7 8 + 1 ) = 1/ 6 ( 7 8 − 1 ) ( 7 8 + 1 ) = 1/ 6 ( 7 16 − 1 ) Vì 7 16 − 1 > 1/ 6 ( 7 16 − 1 ) nên 7 16 − 1 > 8 ( 7 8 + 1 ) ( 7 4 + 1 ) ( 7 2 + 1 )\)
Bài 6:
a: \(2003\cdot2005\)
\(=\left(2004-1\right)\left(2004+1\right)\)
\(=2004^2-1< 2004^2\)
b: \(8\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\)
\(=\left(7^4-1\right)\left(7^4+1\right)\left(7^8+1\right)\)
\(=\left(7^8-1\right)\left(7^8+1\right)\)
\(=7^{16}-1\)
