Question

So sánh:

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

p/s: có cách làm nhé

Answer 1

Ta có: \(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^8-1\right)\left(2^8+1\right)\)
\(B=2^{16}-1\) < A
Vậy A > B

Answer 2

Ta có:

\(A=2^{16}\)

\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(B=\left(2+1\right)\left(2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(B=\left(2^8-1\right)\left(2^8+1\right)\)

\(B=2^{16}-1< 2^{16}\)

Vậy A > B

Answer 3

Ta có : \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)

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

Vì \(2^{16}>2^{16}-1\Rightarrow A>B\)