Lời giải:
Sử dụng công thức \((a-b)(a+b)=a^2-b^2\)
\(A=3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)\)
\(A=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)\)
\(=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)\)
\(=(2^8-1)(2^8+1)(2^{16}+1)\)
\(=(2^{16}-1)(2^{16}+1)=2^{32}-1\)
3(22+1)(24+1)(28+1)(216+1)
=\(\dfrac{3}{2^2-1}\)(22 -1)(22+1)(24+1)(28+1)(216+1)
=\(\dfrac{3}{4-1}\)(24-1)(24+1)(28+1)(216+1)
=\(\dfrac{3}{3}\)(28-1)(28+1)(216+1)
=(216-1)(216+1)
=232-1
3(22+1)(24+1)(28+1)(216+1)
=\(\dfrac{3}{2^2-1}\)(22-1)(22+1)(24+1)(28+1)(216+1)
=(24-1)(24+1)(28+1)(216+1)
=(28-1)(28+1)(216+1)
=(216-1)(216+1)
=232-1
!!!Chúc học tốt!!!