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28 tháng 6 2018

có trong câu hỏi tương tự đấy bạn

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

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

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

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

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

23 tháng 9 2021

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

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

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

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

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

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

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

 

11 tháng 8 2017

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

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

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

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

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

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

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

11 tháng 8 2017

Đinh Đức Hùng lấy 52 -1 ở đâu đấy

19 tháng 8 2020

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

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

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

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

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

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

     \(2A=\left(5^{16}\right)^2-1^2\)

     \(2A=5^{32}-1\)

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

1 tháng 7 2016

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

\(\Rightarrow2Q=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

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

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

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

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

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

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

18 tháng 6 2016

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 \(A=\frac{5^{32}-1}{2}\)

18 tháng 6 2016

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

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

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

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

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

15 tháng 6 2016

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

15 tháng 6 2016

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

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

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

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

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

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

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

16 tháng 8 2016

Theo đầu bài ta có:
\(12\cdot\left(5^2+1\right)\cdot\left(5^4+1\right)\cdot\left(5^8+1\right)\cdot\left(5^{16}+1\right)\)
\(=\frac{24}{2}\cdot\left(5^2+1\right)\cdot\left(5^4+1\right)\cdot\left(5^8+1\right)\cdot\left(5^{16}+1\right)\)
\(=\frac{\left(5^2-1\right)\cdot\left(5^2+1\right)\cdot\left(5^4+1\right)\cdot\left(5^8+1\right)\cdot\left(5^{16}+1\right)}{2}\)
\(=\frac{\left(5^4-1\right)\cdot\left(5^4+1\right)\cdot\left(5^8+1\right)\cdot\left(5^{16}+1\right)}{2}\)
\(=\frac{\left(5^8-1\right)\cdot\left(5^8+1\right)\cdot\left(5^{16}+1\right)}{2}\)
\(=\frac{\left(5^{16}-1\right)\cdot\left(5^{16}+1\right)}{2}\)
\(=\frac{5^{32}-1}{2}\)