Question

So sánh: A = (3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1) với B = 3³² - 1

Answer 1

\(A=\left(3+1\right)\left(3^2+1\right)....\left(3^{16}+1\right)\)

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

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

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

\(2A=\left(3^{32}-1\right)=B\Rightarrow B>A\)

Answer 2

Theo bài ra ta có:

\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ \Rightarrow2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ \Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ \Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ \Rightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ \Rightarrow2A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ \Rightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\\ \Rightarrow2A=3^{32}-1\\ \Rightarrow A=\dfrac{3^{32}-1}{2}\) ta thấy : \(\dfrac{3^{32}-1}{2}< 3^{32}-1\\ \)

=> A < B

vậy A < B

Answer 3

\(\text{Ta có : }A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ A=\dfrac{2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\\ A=\dfrac{\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\\ A=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\\ A=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\\ A=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\\ A=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2}\\ A=\dfrac{3^{32}-1}{2}\\ Do\text{ }\dfrac{3^{32}-1}{2}< 3^{32}-1\\ nên\text{ }\Rightarrow A< B\)

Vậy \(A< B\)