K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
20 tháng 12 2016
3 (22 + 1)(24 + 1)(28 + 1)(216 + 1)
=(22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
=(24 - 1)(24 + 1)(28 + 1)(216 + 1)
=(28 - 1)(28 + 1)(216 + 1)
=(216 - 1)(216 + 1)
=232 -1
NN
8 tháng 7 2019
3. ( 22 + 1 ).( 24 + 1 ).( 28 + 1 )......( 264 + 1 ) + 1
= ( 22 - 1 ).( 22 + 1 ).( 24 + 1 ).( 28 + 1 )....( 264 + 1 ) + 1
= ( 24 - 1 ).( 24 + 1 ).( 28 + 1 )......( 264 + 1 ) + 1
= ( 28 + 1 ).....( 264 + 1 ) + 1
= ( 264 - 1 ).( 264 + 1 ) + 1
= 2128 - 1 + 1
= 2128
NN
8 tháng 7 2019
8.( 32 + 1 ).( 34 + 1 ).( 38 + 1 )....( 3128 + 1 ) + 1
= ( 32 - 1 ).( 32 + 1 ).( 34 + 1 ).( 38 + 1 )....( 3128 + 1 ) + 1
= ( 34 - 1 ).( 34 + 1 ).( 38 + 1 )....( 3128 + 1 ) + 1
= ( 38 - 1 ).( 38 + 1 )....( 3128 + 1 ) + 1
= ( 316 - 1 )......( 3128 + 1 ) + 1
= ( 3128 - 1 ).( 3128 + 1 ) + 1
= 3256 - 1 + 1
= 3256
SK
6 tháng 12 2016
3(2^2+1)(2^4+1)(2^8+1)(2^16+1)
=(2^2-1)(2^2+1)(2^4+1).......
=(2^2-1)(2^2+1) là hằng đẳng thức và = (2^4-1)
tương tự cứ như thế kết quả biểu thức là 2^32-1
6 tháng 12 2016
\(A=\left(....\right)\)
3=4-1=(2^2-1)
A.=(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
KL
A=\(2^{32}-1\)
7 tháng 10 2017
\(100^2-99^2+98^2-97^2+...+2^2-1\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+....+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+....+\left(2-1\right)\left(2+1\right)\)
\(=1.199+1.195+...+1.3\)
\(=199+195+....+3\)
\(=\left[\left(\dfrac{199-3}{4}\right)+1\right]:2.\left(199+3\right)=5050\)
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{3^{32}-1}{2}\)
\(3\left(2^2+1\right)\left(2^4+1\right).....\left(2^{64}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{64}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right).....\left(2^{64}+1\right)\)
\(=\left(2^8-1\right)......\left(2^{64}+1\right)=2^{128}-1\)
KL
1
HH
26 tháng 6 2018
Giải:
a) \(M=\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(\Leftrightarrow3M=2^{32}-1\)
\(\Leftrightarrow M=\dfrac{2^{32}-1}{3}\)
Vậy ...
b) \(N=16\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=48\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=\left(7^4-1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=\left(7^8-1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=\left(7^{16}-1\right)\left(7^{16}+1\right)\)
\(\Leftrightarrow3N=7^{32}-1\)
\(\Leftrightarrow N=\dfrac{7^{32}-1}{3}\)
Vậy ...
NT
1
DN
24 tháng 10 2018
\(A=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Rightarrow A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Rightarrow A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Rightarrow A=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Rightarrow A=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Rightarrow A=\left(2^{32}-1\right)\left(2^{32}+1\right)\)
\(\Rightarrow A=2^{64}-1\)
Vậy \(A=2^{64}-1\)
NT
1
5 tháng 11 2018
\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(A=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(A=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(A=\left(2^{32}-1\right)\left(2^{32}+1\right)\)
\(A=2^{64}-1\)
\(3.\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right)\)
\(=\left(2^2-1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right)\)
\(=\left(2^4-1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right)\)
\(=\left(2^8-1\right).\left(2^8+1\right).\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right).\left(2^{16}+1\right)\)
\(=2^{32}-1.\)
Chúc bạn học tốt!