Câu hỏi của Nguyễn Thị Tuyết Mai

Question

Bài 1 so sánh

A=2012×2014 và B=2013^2

A=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1) và B=3^32-1

A=2017^2-17^2 vàB= 2000^2

Pham Minh Hoang · Accepted Answer

A=2012x2014=2012x(2012+2)=2012^2+4024

B=2013^2=(2012+1)^2=2012^2+2x2012+1=2012^2+2025

=>A

chúc bạn học tốt~~~

Câu trả lời:

A=2012x2014=2012x(2012+2)=2012^2+4024

B=2013^2=(2012+1)^2=2012^2+2x2012+1=2012^2+2025

=>A

chúc bạn học tốt~~~

Phùng Minh Quân · Answer

Bài 1 :

$a)$$A=2012.2014=\left(2013-1\right)\left(2013+1\right)=2013^2-1< 2013^2=B$

Vậy $A< B$

$b)$$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)$

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

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

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

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

$2A=\left(3^{16}-1\right)\left(3^{16}+1\right)$

$2A=3^{32}-1$

$A=\frac{3^{32}-1}{2}< 3^{32}-1=B$

$c)$$A=2017^2-17^2=\left(2017-17\right)\left(2017+17\right)=2000.2034>2000.2000=2000^2=B$

Vậy $A>B$