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Duy Lê
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nguyễn thị huyền anh
28 tháng 6 2018 lúc 21:59

có trong câu hỏi tương tự đấy bạn

Phan An
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Nguyễn Lê Phước Thịnh
23 tháng 9 2021 lúc 21:02

\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(=\dfrac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{2}\)

\(=\dfrac{5^{32}-1}{2}\)

Minh Hiếu
23 tháng 9 2021 lúc 21:04

\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{32}+1\right)\)

\(2P\)\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(2P=\)\(\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(2P=\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(2P=5^{32}-1\)

\(P=\dfrac{5^{32}-1}{2}\)

 

Võ Anh Quân
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Thanh Hằng Nguyễn
10 tháng 9 2017 lúc 13:25

Đặt :

\(A=\)\(12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(\Leftrightarrow2A=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^{16}+1\right)\left(5^{16}-1\right)\)

\(=5^{32}-1\)

\(\Leftrightarrow A=\frac{5^{32}-1}{2}\)

Trương Tùng
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le van thang
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Không Tên
30 tháng 7 2018 lúc 20:44

\(P=12.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(=\frac{5^{32}-1}{2}\)

Trần Phượng
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Vũ Minh Tuấn
19 tháng 8 2020 lúc 18:52

Đặt \(A=12.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)

\(\Rightarrow2A=24.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)

     \(2A=\left(5^2-1\right).\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)

     \(2A=\left(5^4-1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)

     \(2A=\left(5^8-1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)

     \(2A=\left(5^{16}-1\right).\left(5^{16}+1\right)\)

     \(2A=\left(5^{16}\right)^2-1^2\)

     \(2A=5^{32}-1\)

\(\Rightarrow A=\frac{5^{32}-1}{2}.\)

Khách vãng lai đã xóa
Ayakashi
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MrKakashi
17 tháng 7 2016 lúc 14:00

\(A=\frac{1}{2}\left(5^2-1\right)\) \(\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\) 

A= 1/2(5^4-1)(5^4+1)(5^8+1)(5^16+1)

Làm tương tự còn:

A=1/2(5^16-1)(5^16+1)=1/2(5^32-1)

pham thi thu thao
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Không Tên
3 tháng 7 2018 lúc 16:12

\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(=\frac{5^{32}-1}{2}\)

Lê Hà Vy
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Đinh Tuấn Việt
18 tháng 6 2016 lúc 17:02

Đặt \(A=12.\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(2A=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(=5^{32}-1\)

Vậy \(A=\frac{5^{32}-1}{2}\)

Nguyễn Thị Anh
18 tháng 6 2016 lúc 17:02

\(\frac{12.\left(5^2+1\right)\left(5^2-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{5^2-1}\)

=\(\frac{12.\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{24}\)

=\(\frac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

=\(\frac{\left(5^{16}-1\right)\left(5^{16+1}\right)}{2}\)

=\(\frac{5^{32}-1}{2}\)