Question

Bài 1: So sánh 

A= (2+1)*(2^2+1)*(2^4+1)*(2^16+1)

B= (2^23-1)

Bài 2: Cho x-y=5 và y*x=14

Tính A= x^2+y^2

B= (x+y)^2

 

Answer 1

1/

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

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

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)

Vì 2³² > 2²³ => 2³²-1>2²³-1 hay A>B

2/

\(A=x^2+y^2=\left(x-y\right)^2+2xy=5^2+2.14=25+28=53\)

\(B=\left(x+y\right)^2=\left(x-y\right)^2+4xy=5^2+4.14=25+56=81\)