Bạn chưa đăng nhập. Vui lòng đăng nhập để hỏi bài

Những câu hỏi liên quan
Minh Châu
Xem chi tiết
Thu Thao
11 tháng 12 2020 lúc 20:30

\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

I
Xem chi tiết
Nguyễn Lê Phước Thịnh
14 tháng 7 2021 lúc 21:13

a) Ta có: \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

Pham Trong Bach
Xem chi tiết
Cao Minh Tâm
29 tháng 3 2019 lúc 4:38

Ta có

N   =   ( 2   +   1 ) ( 2 2   +   1 ) ( 2 4   +   1 ) ( 2 8   +   1 ) ( 2 16   +   1 )     ( 2 16   +   1 )   =   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 16 2 − 1 = 2 32 − 1 M à   2 32 − 1 > 2 32 ⇒   N < M

Đáp án cần chọn là: A

Uyen Hoang
Xem chi tiết
Lê Thị Thục Hiền
9 tháng 7 2021 lúc 16:30

\(A=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

✿✿❑ĐạT̐®ŋɢย❐✿✿
9 tháng 7 2021 lúc 16:31

Đặt : \(P=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

3(22+1)(24+1)(28+1)(216+1

=\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

=\(\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

=\(\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

=\(\left(2^{16}-1\right)\left(2^{16}+1\right)\)

=\(2^{32}-1\)

Phạm Nguyễn Ngọc Mai
Xem chi tiết
Yeutoanhoc
19 tháng 6 2021 lúc 14:03

`A=(2-1)(2+1)(2^2+1)...(2^16+1)`

`=(2^2-1)(2^2+1)....(2^16+1)`

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

`=2^32-1<2^32`

`=>A<B`

Lý Bá Đức Thịnh
Xem chi tiết
Duc Nguyen
10 tháng 12 2023 lúc 14:51

1,

Đặt \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(\left(2-1\right)A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(1A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(A=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(A=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(A=2^{32}-1\)

Vậy \(A=2^{32}-1\)

2, \(x^2-6x=-9\)

\(x^2-6x+9=0\)

\(\left(x-3\right)^2=0\)

\(x-3=0\)

\(x=3\)

Vậy \(x=3\)

Đã Ẩn
Xem chi tiết
thangdeptrai
Xem chi tiết
Nguyễn Lê Phước Thịnh
12 tháng 7 2023 lúc 9:41

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

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

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

=(2^16-1)(2^16+1)+1

=2^32-1+1

=2^32

HT.Phong (9A5)
12 tháng 7 2023 lúc 9:42

\(3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{16}+1\right)+1\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{16}+1\right)+1\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)+1\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)+1\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)+1\)

\(=\left(2^{32}-1\right)+1\)

\(=2^{32}\)

Đức Anh Vũ
Xem chi tiết