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Ta có: \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot...\cdot\left(3^{64}+1\right)\)

\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot...\cdot\left(3^{64}+1\right)}{2}\)

\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\cdot...\cdot\left(3^{64}+1\right)}{2}\)

\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)}{2}\)

\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)}{2}\)

\(=\dfrac{\left(3^{32}-1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)}{2}\)

\(=\dfrac{\left(3^{64}-1\right)\left(3^{64}+1\right)}{2}\)

\(=\dfrac{3^{128}-1}{2}\)

Phong Thần

bạn để ý quan hệ giữa các phần tử trong ngoặc
*thực ra toi nhân ngẫu nhiên số 2 vào nếu ko đẹp thì sẽ là nhân 4

3: =(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

4:

D=(4^4-1)(4^4+1)(4^8+1)*....*(4^64+1)

=(4^8-1)(4^8+1)*...*(4^64+1)

=...

=4^128-1

5: =(5^2-1)(5^2+1)(5^4+1)*...*(5^128+1)+(5^256-1)

=(5^4-1)(5^4+1)*...*(5^128+1)+5^256-1

=5^256-1+5^256-1

=2*5^256-2

7 tháng 7 2023

thsu là rất ngưỡng mộ anh ạ 🥹 em mấy lần off vì quá nhác nhưng lần nào ngoi lại lên cũng thấy anh cày chăm chỉ quá tr 😭

24 tháng 8 2018

nhiều thế, đăng ít một thôi bạn

24 tháng 8 2018

a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)

24 tháng 6 2016

Giúp vs nhé mk sẽ dùng các nick phụ tới tấp cho mí bạn

14 tháng 7 2016

a) (2+1)(2^2+1)(2^4+1)...(2^32+1)-2^64

=(2+1)(2-1)(2^2+1)(2^4+1)...(2^32+1)-2^64

=(2^2-1)(2^2+1)(2^4+1)...(2^32+1)-2^64

=(2^4-1)(2^4+1)....(2^32+1)-2^64

=......

=(2^32-1)(2^32+1)-2^64

=2^64-1-2^64=-1

b)Đặt A=(5+3)(5^2+3^2)(5^4+3^4)...(5^64+3^64)+(5^128-3^128)/2

đặt B=(5+3)(5^2+3^2)(5^4+3^4)...(5^64+3^64)

\(2B=\left(5-3\right)\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)

\(2B=\left(5^2-3^2\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)

\(2B=\left(5^4-3^4\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)

\(2B=.......\)

2B=(5^64-3^64)(5^64+3^64)

2B=5^128-3^128

B=(5^128-3^128)/2 (thế vào đề bài)

=> A=B+(5^128-3^128)/2=(5^128-3^128)/2+(5^128-3^128)/2=\(\frac{2\left(5^{128}-3^{128}\right)}{2}=\left(5^{128}-3^{128}\right)\)

14 tháng 7 2016

a) A = ( 2-1)(2+1)(22+1)...(232+1)-264

         =(22-1)(22+1)(24+1)... -264

       =....

       =264-1-264=1

câu b tương tự nhá

5 tháng 11 2017

a, \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}=2^{64}-1-2^{64}=-1\)

b,\(B=\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)+\dfrac{5^{128}-3^{128}}{2}\)

\(=\dfrac{\left(5-3\right)\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)}{2}+\dfrac{5^{128}-3^{128}}{2}\)\(=\dfrac{\left(5^2-3^2\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)+5^{128}-3^{128}}{2}\)

\(=\dfrac{\left(5^{64}-3^{64}\right)\left(5^{64}+3^{64}\right)+5^{128}-3^{128}}{2}=\dfrac{2.5^{128}}{2}=5^{128}\)

1: A=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)

=(3^4-1)(3^4+1)(3^8+1)(3^16+1)

=(3^8-1)(3^8+1)(3^16+1)

=(3^16-1)(3^16+1)

=3^32-1

2: B=(1-3^2)(1+3^2)*...*(1+3^16)

=(1-3^4)(1+3^4)(1+3^8)(1+3^16)

=1-3^32

7 tháng 7 2023

1

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

 

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

8 tháng 3 2020

Ta có \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)

\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)

\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)

\(=\left(3^4-1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)

\(=\left(3^{64}-1\right)\left(3^{64}+1\right)=\left(3^{128}-1\right)\)

\(\Rightarrow A=\frac{3^{128}-1}{2}\)