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TN
2
12 tháng 7 2016
Đặt \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\).Ta có :
\(=>\left(3-1\right)A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=>2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=>2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
...............................................................................
Cuối cùng \(=>2A=3^{64}-1\).
\(=>A=\frac{3^{64}-1}{2}\)
12 tháng 7 2016
Đặt \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=...........................................\)
\(=\left(3^{32}-1\right)\left(3^{32}+1\right)=3^{64}-1\)
\(\Rightarrow A=\frac{3^{64}-1}{2}\)
5 tháng 6 2017
Đặt biểu thức đã cho là A.
Ta có: 2A = (3 - 1) * (3 + 1) * (3^2 + 1) * .... * (3^64 + 1)
= (3^2 - 1) * (3^2 + 1) * ... * (3^64 + 1) (hằng đẳng thức a^2 - b^ 2 = (a+b)(a-b))
Rút gọn triệt tiêu ta được 2A=3^64 - 1
=> A = (3^64 - 1)/2
TN
1
NN
1
23 tháng 7 2019
\(8.\left(3^2+1\right).\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)-3^{32}\)
\(=\left(3^2-1\right).\left(3^2+1\right).\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)-3^{32}\)
\(=\left(3^4-1\right).\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)-3^{32}\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)-3^{32}\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)-3^{32}=3^{32}-1-3^{32}=-1\)
22 tháng 7 2016
\(B=\left(3-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(\left(3+1\right)B=\left(3+1\right)\left(3-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(\left(3+1\right)B=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(\left(3+1\right)B=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(\left(3+1\right)B=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(\left(3+1\right)B=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(\left(3+1\right)B=\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(\left(3+1\right)B=3^{64}-1\)
\(B=\frac{3^{64}-1}{4}\)
Chúc bạn làm bài tốt
HN
1
29 tháng 10 2021
\(=3\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{8}\)
\(=\dfrac{3\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{8}\)
\(=\dfrac{3\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{8}\)
\(=\dfrac{3\left(3^{16}-1\right)\left(3^{16}+1\right)}{8}\)
\(=\dfrac{3\left(3^{32}-1\right)}{8}\)
P
3
VA
6 tháng 10 2016
Ta có: 3 + 1 = (3^2 - 1)/(3 - 1)
3^2 + 1 = (3^4 - 1)/(3^2 - 1)
3^4 + 1 = (3^8 - 1)/(3^4 - 1)
3^8 + 1 = (3^16 - 1)/(3^8 - 1)
3^16 + 1 = (3^32 - 1)/(3^16 - 1)
3^32 + 1 = (3^64 - 1)/(3^32 - 1)
(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
=(3^2 - 1)/(3 - 1).(3^4 - 1)/(3^2 - 1).(3^8 - 1)/(3^4 - 1).(3^32 - 1)/(3^16 - 1).(3^64 - 1)/(3^32 - 1)
=(3^64 - 1)/(3 - 1)
=(3^64 - 1)/2
7 tháng 10 2016
Đặt biểu thức đó là A
(3-1) A= (3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1) (3^32+1)
2 A= (3^2-1)(3^2+1)(3^4+1)..............................................
2A = (3^4-1)(3^4+1)(3^8+1) ............................
2A= (3^8-1)(3^8+1)(3^16+1) .............
2A = (3^16-10(3^16+1)(3^32+1)
2A = (3^32-1)(3^32+1)
2A= 3^64-1
A= (3^64-1) / 2
Đặt \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
Ta có: \(2.A=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)=\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(=3^{64}-1=>A=\frac{3^{64}-1}{2}\)