Question

So sánh hai số bằng cách vận dụng hằng đẳng thức :

a, A = 1999 . 2001 và B = \(2000^2\)

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

Answer 1

a)A=\(1999.2001=\left(2000-1\right)\left(2000+1\right)=2000^2-1\)

Vậy A < B

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

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

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

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

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

Vậy B < A

Answer 2

a) Ta có: \(A=1999.2001=\left(2000-1\right)\left(2000+1\right)\)

\(=2000^2-1^2< 2000^2\)

Vậy A < B.

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

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

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

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

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

Vậy A > B.

Answer 3

\(A=1999.2001=\left(2000+1\right)\left(2000-1\right)=2000^2-1\)

\(A< B\)

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

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

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

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

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

\(B< A\)