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Thương Nguyễn
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Tài Nguyễn Tuấn
12 tháng 7 2016 lúc 20:55

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 \(=>2A=3^{64}-1\).

\(=>A=\frac{3^{64}-1}{2}\)

Hoàng Lê Bảo Ngọc
12 tháng 7 2016 lúc 21:31

Đặt \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=...........................................\)

\(=\left(3^{32}-1\right)\left(3^{32}+1\right)=3^{64}-1\)

\(\Rightarrow A=\frac{3^{64}-1}{2}\)

Karry Wang
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Trương Quang Khải
5 tháng 6 2017 lúc 17:33

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

Đức Anh Lê
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Trần Nguyễn Phương Thủy
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BLACK WHITE
4 tháng 11 2020 lúc 21:32

goi y nha A=1/2.(3^2-1)(3^2+1)....(3^32+1)

Khách vãng lai đã xóa
Nguyễn Ngọc Thùy Nga
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headsot96
23 tháng 7 2019 lúc 15:35

\(8.\left(3^2+1\right).\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)-3^{32}\)

\(=\left(3^2-1\right).\left(3^2+1\right).\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)-3^{32}\)

\(=\left(3^4-1\right).\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)-3^{32}\)

\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)-3^{32}\)

\(=\left(3^{16}-1\right)\left(3^{16}+1\right)-3^{32}=3^{32}-1-3^{32}=-1\)

Nghi Baekkie
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Lê Nguyên Hạo
22 tháng 7 2016 lúc 6:25

3 + 1 hay 3 - 1 z

Trần Thu Uyên
22 tháng 7 2016 lúc 6:41

\(B=\left(3-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(\left(3+1\right)B=\left(3+1\right)\left(3-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(\left(3+1\right)B=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(\left(3+1\right)B=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(\left(3+1\right)B=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(\left(3+1\right)B=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(\left(3+1\right)B=\left(3^{32}-1\right)\left(3^{32}+1\right)\)

\(\left(3+1\right)B=3^{64}-1\)

\(B=\frac{3^{64}-1}{4}\)

Chúc bạn làm bài tốt

Hoài Nam
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Nguyễn Lê Phước Thịnh
29 tháng 10 2021 lúc 20:51

\(=3\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{8}\)

\(=\dfrac{3\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{8}\)

\(=\dfrac{3\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{8}\)

\(=\dfrac{3\left(3^{16}-1\right)\left(3^{16}+1\right)}{8}\)

\(=\dfrac{3\left(3^{32}-1\right)}{8}\)

phuong
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Vũ Anh Quân
6 tháng 10 2016 lúc 21:49

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

Nguyễn Thị Hằng
7 tháng 10 2016 lúc 20:59

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

T.Ps
12 tháng 7 2019 lúc 16:35

#)Giải :

\(=\frac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{8}\)

\(=\frac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{8}\)

\(=\frac{\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{8}\)

\(=\frac{\left(3^{32}-1\right)\left(3^{32}+1\right)}{8}\)

\(=\frac{3^{64}-1}{8}\)

Sakura Kinomoto
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