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

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

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

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

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

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

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

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

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

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

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

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

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

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

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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)

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Hoàng Thị Cẩm Hà
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Nguyễn Thị Kim Phương
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KWS
24 tháng 8 2018 lúc 14:06

2p=24(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(5^4-1)(5^4+1)(5^8+1)(5^16+1)
=(5^8-1)(5^8+1)(5^16+1)
=(5^16-1)(5^16+1)
=5^32-1
~> p=5^32-1/2

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Linh
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Thảo Lê Thị
30 tháng 6 2016 lúc 8:13

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

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

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

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

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

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Nguyễn Đức Tiến
30 tháng 6 2016 lúc 8:20

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

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

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

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

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

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

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

Vậy \(P=\frac{5^{32}-1}{2}\)

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Sách Giáo Khoa
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T.Thùy Ninh
22 tháng 6 2017 lúc 19:31

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

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

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

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Trần Quốc Lộc
18 tháng 8 2017 lúc 17:19

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

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

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

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

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

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

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

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Nhật Minh
20 tháng 5 2017 lúc 18:22

Viết 12 = 1/2 . ( 52 -1 ) sau đó áp dụng HĐT thứ 3 nhé .

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

Đặt :

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

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

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

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

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

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

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

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

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Phương Kuro
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Minh Anh
19 tháng 9 2016 lúc 18:02

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

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

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

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

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

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

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