Đáp án+Giải thích các bước giải:
Ta có:
`A=4(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)`
`=>2A=2.4(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)`
`=8.(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)`
`=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)`
`=(3^4-1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)`
`=(3^8-1)(3^8+1)(3^16+1)(3^32+1)`
`=(3^16-1)(3^16+1)(3^32+1)`
`=(3^32-1)(3^32+1)`
`=3^64-1`
`=>A={3^64-1}/2`
Vì `{3^64-1}/2<3^64-1`
hay `4(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)<3^64-1`
`=>A<B`
Vậy `A<B`
