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![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt :
\(A=\)\(12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow2A=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^{16}+1\right)\left(5^{16}-1\right)\)
\(=5^{32}-1\)
\(\Leftrightarrow A=\frac{5^{32}-1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(=\frac{5^{32}-1}{2}\)
Rút gọn biểu thức
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(=5^{32}-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^{32}-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{32}-1\right)\)
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 \(P=\frac{5^{32}-1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+16\right)\)\(=\dfrac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)\(=\dfrac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\cdot\left(5^8+1\right)\cdot\left(5^{16}+1\right)\)\(=\dfrac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\dfrac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(=\dfrac{1}{2}\left(5^{32}-1\right)\)
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\dfrac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\dfrac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\dfrac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\dfrac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(P=\dfrac{1}{2}\left(5^{32}-1\right)\)
\(P=\dfrac{5^{32}-1}{2}\)
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
Ta có :
\(2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(5^{32}-1\)
\(\Rightarrow P=\frac{5^{32}-1}{2}\)
ta có : P=\(12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)=\frac{12\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{\left(5^2-1\right)}\)
=> P=\(\frac{5^{32}-1}{2}\)