Câu hỏi của Trần Thị Tuyết Ngân

Question

Cho A = (3 + 1 ).(3^2 + 1 ).(3^4 + 1 ).(3^8 + 1 ).(3^16 + 1 )

B = (3^32 - 1 ) .Khi đó, B = k.A

Vậy k =?

Đỗ Thanh Hải · Accepted Answer

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

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

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

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

$\Rightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)$

$\Rightarrow2A=3^{32}-1$

$\Rightarrow A=\dfrac{3^{32}-1}{2}$

$\Rightarrow A.2=B=3^{32}-1$

$\Rightarrow k=2$

Vậy $k=2$ là giá trị cần tìmCâu trả lời: $A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)$