Question

Answer 1

Lời giải:

Áp dụng HĐT $(a-b)(a+b)=a^2-b^2$

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

$=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)$

$=(3^8-1)(3^8+1)(3^{16}+1)$

$=(3^{16}-1)(3^{16}+1)=3^{32}-1$
2.

$-B=(3-1)(3+1)(3^2+1)(3^4+1)...(3^{16}+1)$

$-B=(3^2-1)(3^2+1)(3^{14}+1)...(3^{16}+1)=A=3^{32}-1$

$\Rightarrow B=1-3^{32}$

3.

$C=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^{16}+1)$

$=(5^4-1)(5^4+1)(5^8+1)(5^{16}+1)$

$=(5^8-1)(5^8+1)(5^{16}+1)$

$=(5^{16}-1)(5^{16}+1)=5^{32}-1$

4. Tương tự 3 bài trước:

$D=(4^2-1)(4^2+1)(4^4+1)...(4^{64}+1)=4^{128}-1$

5. $E=5^{256}-1$