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17 tháng 7 2016

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

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)

3 tháng 7 2018

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

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

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

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

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

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

25 tháng 6 2016

dặt tổng là P

P= 12.(52+1)(54+1)(58+1)(516+1)

=>2P=24.(52+1)(54+1)(58+1)(516+1)

=(52-1)(52+1)(54+1)(58+1)(516+1)

=(54-1)(58+1)(516+1)

=(58-1)(58+1)(516+1)

=(516-1)(516-1)

=532-1

=>(532-1 ):2

25 tháng 6 2016

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

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

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

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

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

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

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

24 tháng 8 2018

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

30 tháng 6 2016

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

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

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

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

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

30 tháng 6 2016

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 \(P=\frac{5^{32}-1}{2}\)

22 tháng 6 2017

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

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

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

18 tháng 8 2017

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

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

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

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

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

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

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

10 tháng 9 2017

Đặt :

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

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

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

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

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

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

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

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

6 tháng 8 2016

\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{15}+1\right)\)

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

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

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

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

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

 

6 tháng 8 2016

a)

 Ta có

a chia 5 dư 4

=> a=5k+4 ( k là số tự nhiên )

\(\Rightarrow a^2=\left(5k+4\right)^2=25k^2+40k+16\)

Vì 25k^2 chia hết cho 5

    40k chia hết cho 5

    16 chia 5 dư 1

=> đpcm

2) Ta có

\(12=\frac{5^2-1}{2}\)

Thay vào biểu thức ta có

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

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

\(\Rightarrow P=\frac{\left[\left(5^4\right)^2-1^2\right]\left[\left(5^4\right)^2+1^2\right]}{2}\)

\(\Rightarrow P=\frac{5^{16}-1}{2}\)

3)

\(\left(a+b+c\right)^3=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)

\(=a^3+b^3+c^2+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ca+cb+c^2\right)\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

 

19 tháng 9 2016

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

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

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

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

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

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