Ruts gon BT A=3(2^2+1)(2^4+1)(2^8)(2^16+1)

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Ruts gon BT A=3(2^2+1)(2^4+1)(2^8)(2^16+1)

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Iris 4 tháng 2021-08-25T07:55:48+00:00 1 Answers 4 views 0

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    2021-08-25T07:57:40+00:00

    Đáp án:

     A=2^32-1

    Giải thích các bước giải:

    A=3(2^2+1)(2^4+1)(2^8+1)(2^16+1)

    A=3(2^2+1)(2^4+1)(2^8+1)(2^16+1)

       =(2^2-1).(2^2+1)(2^4+1)(2^8+1)(2^16+1)

       =(2^4-1).(2^4+1)(2^8+1)(2^16+1)

       =(2^8-1).(2^8+1)(2^16+1)

       =(2^16-1).(2^16+1)

       =2^32-1

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