Đặt:
`A=6(7^2+1)(7^4+1)(7^8+1)(7^16+1)`
`B=7^32/8`
Ta có:
`+)A=6(7^2+1)(7^4+1)(7^8+1)(7^16+1)`
`→8A=8.6(7^2+1)(7^4+1)(7^8+1)(7^16+1)`
`8A=48(7^2+1)(7^4+1)(7^8+1)(7^16+1)`
`8A=(7^2-1)(7^2+1)(7^4+1)(7^8+1)(7^16+1)`
`8A=(7^4-1)(7^4+1)(7^8+1)(7^16+1)`
`8A=(7^8-1)(7^8+1)(7^16+1)`
`8A=(7^16-1)(7^16+1)`
`8A=7^32-1`
`+)B=7^32/8`
`→8B=8. 7^32/8`
`8B=7^32`
Do `8A=7^32-1<8B=7^32` nên `A<B`
hay `6(7^2+1)(7^4+1)(7^8+1)(7^16+1)<7^32/8`
Vậy `6(7^2+1)(7^4+1)(7^8+1)(7^16+1)<7^32/8`
