rút gọn biểu thức 3(2^2+1)(2^4+1)(2^8+1)(2^16+1) 05/12/2021 Bởi Hadley rút gọn biểu thức 3(2^2+1)(2^4+1)(2^8+1)(2^16+1)
Đáp án: $ 2^{32} – 1$ Giải thích các bước giải: $\begin{array}{l}\quad 3(2^2 + 1)(2^4 + 1)(2^8+1)(2^{16}+1)\\ = (2-1)(2+1)(2^2 + 1)(2^4 + 1)(2^8+1)(2^{16}+1)\\ = (2^2-1)(2^2 + 1)(2^4 + 1)(2^8+1)(2^{16}+1)\\ = (2^4-1)(2^4 + 1)(2^8+1)(2^{16}+1)\\ = (2^8-1)(2^8+1)(2^{16}+1)\\ = (2^{16}-1)(2^{16}+1)\\ = 2^{32} – 1 \end{array}$ Bình luận
Đáp án: `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` Bình luận
Đáp án:
$ 2^{32} – 1$
Giải thích các bước giải:
$\begin{array}{l}\quad 3(2^2 + 1)(2^4 + 1)(2^8+1)(2^{16}+1)\\ = (2-1)(2+1)(2^2 + 1)(2^4 + 1)(2^8+1)(2^{16}+1)\\ = (2^2-1)(2^2 + 1)(2^4 + 1)(2^8+1)(2^{16}+1)\\ = (2^4-1)(2^4 + 1)(2^8+1)(2^{16}+1)\\ = (2^8-1)(2^8+1)(2^{16}+1)\\ = (2^{16}-1)(2^{16}+1)\\ = 2^{32} – 1 \end{array}$
Đáp án:
`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`