51问答网 > -(3+1)(3^2+1)(3^4+1)(3^8+1)=？要详细过程，10分钟之内发过来

-(3+1)(3^2+1)(3^4+1)(3^8+1)=？要详细过程，10分钟之内发过来

2024-12-17 06:12:24

推荐回答（4个）

回答1：

你好很高兴收到你的求助:

在前面加上(3-1)配成平方差
-(3+1)(3^2+1)(3^4+1)(3^8+1)
=-1/2×(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1) 一直平方差
=-1/2×(3²-1)(3^2+1)(3^4+1)(3^8+1)
=-1/2×(3^4-1)(3^4+1)(3^8+1)
=-1/2×(3^8-1)(3^8+1)
=-1/2×(3^16-1)
=1/2-(3^16)/2

不懂可追问有帮助请采纳祝你学习进步谢谢

回答2：

-(3+1)(3²+1)(3⁴+1)(^8 +1)
=-(1/2)(3-1)(3+1)(3²+1)(3⁴+1)(^8 +1)
=(-1/2)(3²-1(3²+1)(3⁴+1)(3^8 +1)
=(-1/2)(3⁴-1)3⁴+1)(3^8 +1)
=(-1/2)(3^8-1)(3^8 +1)
=(-1/2)(3^16 -1)
=1/2 - 3^16 /2

回答3：

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

回答4：

-(3+1)(3^2+1)(3^4+1)(3^8+1)
=-4*10*82*(6561+1)
=-40*82*6562
=-215 233 60