DN
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TT
11 tháng 12 2020
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
PN
1
19 tháng 6 2021
`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`
14 tháng 7 2021
a) Ta có: \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
CM
29 tháng 3 2019
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
UH
5
9 tháng 7 2021
\(A=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
9 tháng 7 2021
Đặt : \(P=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
LB
1
DN
10 tháng 12 2023
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 \(x=3\)
14 tháng 3 2020
Bài 2
a. (x-2y)2 =2x-4y
b. (2x^2 +3)2 =4x^2+6
c. (x-2) (x^2+2x+4) = x^3-8 (hằng đẳng thức)
d. (2x-1)3 = 6x-3
Xin lỗi mik chỉ lm ổn bài 2 thôi!
12 tháng 7 2023
=(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
H9
HT.Phong (9A5)
CTVHS
12 tháng 7 2023
\(3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{16}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{16}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)+1\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)+1\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)+1\)
\(=\left(2^{32}-1\right)+1\)
\(=2^{32}\)