Verified answer

(3+1)(3^2+1^2)([3^2]^2+1)([3^4]^2+1)([3^8]^2+1) thanh hang dang thuc roi day ban khai trien ra roi dat nhan tu chung roi rut gon

chuc ban lam tot

A=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)

Nhân cả hai vế với 2 (tức là 3-1) để xuất hiện hằng đaửng thức a^2-b^2=(a+b)(a-b).

A.(3-1) =(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)

A.2=(3^2-1).(3^2+1)(3^4+1)(3^8+1)(3^16+1)

A.2=[(3^2)^2-1].(3^4+1)(3^8+1)(3^16+1)

A.2=(3^8-1).(3^8+1)(3^16+1)

A.2=(3^16-1).(3^16+1)

A.2=3^32-1

Vậy A= (3^32-1)/2

A=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)

(3-1).A =(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)

2A= (3^2-1).(3^2+1)(3^4+1)(3^8+1)(3^16+1)

=(3^4-1).(3^4+1)(3^8+1)(3^16+1) =(3^8-1).(3^8+1)(3^16+1) =(3^16-1).(3^16+1) =3^32-1

A= (3^32-1)/2