Bài `17:`

`a) A = 2008 . 2010 = (2009 - 1)(2009 + 1) = 2009^2 - 1`

Ta có : `-1 < 1`

`to 2009^2 - 1 < 2009^2 + 1`

`to A < B`

`b) 2A = 2(3+1)(3^2+1)(3^4+1)(3^8+1)`

`2A = (3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)`

`2A = (3^2-1)(3^2+1)(3^4+1)(3^8+1)`

`2A = (3^4-1)(3^4+1)(3^8+1)`

`2A = (3^8-1)(3^8 + 1) = 3^16 - 1`

`to A = (3^16 - 1)/2`

Vì `-1 < 0 to 3^16 - 1 < 3^16`

`to (3^16 - 1)/2 < (3^16)/2`

`to A < B`