Question

[Hoàng Nhật Anh]

So sánh

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

Answer 1

[Nguyễn Huệ Lam]

Baì này mình mới làm lúc sáng bạn vào câu hỏi tương tự có đấy

Answer 2

[Phạm Tuấn Đạt]

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

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

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

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

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

\(\Rightarrow2A=3^{32}-1\)

\(\Rightarrow A=\frac{3^{32}-1}{2}< 3^{32}-1=C\)