Question

[Hồ Quỳnh Thơ]

tính và so sánh

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

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

Answer 1

[Không Tên]

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

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

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

                \(.........\)

\(=\frac{1}{2}\left(3^{32}-1\right)\)\(< \)\(3^{32}-1\)\(=\)\(A\)

Vậy  \(B< A\)

Answer 2

[ppcasd]

 A=1.853020189*10 \(^{15}\)

B= 9.265100944*10\(^{15}\)

tự so sánh