Đáp án:
`(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)`
`=(2.(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1))/2`
`=((3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1))/2`
`=((3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1))/2`
`=((3^4-1)(3^4+1)(3^8+1)(3^16+1))/2`
`=((3^8-1)(3^8+1)(3^16+1))/2`
`=((3^16-1)(3^16+1))/2`
`=(3^32-1)/2`
Vì `3^32-1>0`
`<=>2(3^32-1)>3^32-1`
`=>3^32-1>(3^32-1)/2`
Hay `(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)<3^32-1`