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TP
1
VM
19 tháng 8 2020
Đặt \(A=12.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(\Rightarrow2A=24.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(2A=\left(5^2-1\right).\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(2A=\left(5^4-1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(2A=\left(5^8-1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(2A=\left(5^{16}-1\right).\left(5^{16}+1\right)\)
\(2A=\left(5^{16}\right)^2-1^2\)
\(2A=5^{32}-1\)
\(\Rightarrow A=\frac{5^{32}-1}{2}.\)
B
29 tháng 8 2020
Bài làm :
Ta có:
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\frac{24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(P=\frac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(P=\frac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(P=\frac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(P=\frac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{2}\)
\(P=\frac{5^{32}-1}{2}\)
\(\text{Vậy : }P=\frac{5^{32}-1}{2}\)
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
NC
29 tháng 8 2020
Bài làm:
Đặt \(A=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
=> \(2A=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
<=> \(2A=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
<=> \(2A=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
<=> \(2A=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
<=> \(2A=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
<=> \(2A=5^{32}-1\)
=> \(A=\frac{5^{32}-1}{2}\)
28 tháng 5 2018
\(12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\)\(\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\)\(\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)\(=\left(5^{16}-1\right)\left(5^{16}+1\right):2\)
\(=\frac{5^{32}-1}{2}\)
26 tháng 6 2016
\(p=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(p=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(p=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(p=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(p=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(p=\frac{1}{2}\left(5^{32}-1\right)=\frac{5^{32}-1}{2}\)
16 tháng 8 2016
Theo đầu bài ta có:
\(12\cdot\left(5^2+1\right)\cdot\left(5^4+1\right)\cdot\left(5^8+1\right)\cdot\left(5^{16}+1\right)\)
\(=\frac{24}{2}\cdot\left(5^2+1\right)\cdot\left(5^4+1\right)\cdot\left(5^8+1\right)\cdot\left(5^{16}+1\right)\)
\(=\frac{\left(5^2-1\right)\cdot\left(5^2+1\right)\cdot\left(5^4+1\right)\cdot\left(5^8+1\right)\cdot\left(5^{16}+1\right)}{2}\)
\(=\frac{\left(5^4-1\right)\cdot\left(5^4+1\right)\cdot\left(5^8+1\right)\cdot\left(5^{16}+1\right)}{2}\)
\(=\frac{\left(5^8-1\right)\cdot\left(5^8+1\right)\cdot\left(5^{16}+1\right)}{2}\)
\(=\frac{\left(5^{16}-1\right)\cdot\left(5^{16}+1\right)}{2}\)
\(=\frac{5^{32}-1}{2}\)
19 tháng 6 2015
2P = 24.(5^2 + 1 )(5^4 + 1) ... (5^16 + 1)
2P = (5^2 - 1) (5^2 + 1) (5^4 + 1) .. (5^16+1)
2P = (5^4 - 1 )(5^4 + 1 ) (5^8 + 1)
2P = (5^8 - 1 ) (5^8 + 1) (5^16 + 1)
2P = ( 5^ 16 - 1 ) 5^ 16 + 1)
2P = 5^32 - 1
P = (5^32 - 1) : 2
DN
19 tháng 6 2015
\(P=12.\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Rightarrow2P=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow2P=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow2P=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow2P=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow2P=5^{32}-1\)
\(\Leftrightarrow P=\frac{5^{32}-1}{2}\)
Đặt đa thức \(12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\) là A:
\(2A=24.\left(5^2+1\right)\left(5^4+1\right)...\left(5^{16}+1\right)\)
\(2A=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{16}+1\right)\)
\(2A=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\)
\(2A=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2A=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(2A=5^{32}-1\)
\(A=\frac{5^{32}-1}{2}\)