已知M=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/05/29 12:01:16
已知M=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128+1) 求M个位数是几
M=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128+1)
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128+1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128+1)
=(2^8-1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128+1)
=(2^16-1)(2^16+1)(2^32+1)(2^64+1)(2^128+1)
=(2^32-1)(2^32+1)(2^64+1)(2^128+1)
=(2^64-1)(2^64+1)(2^128+1)
=(2^128-1)(2^128+1)
=2^256 - 1
个位数 与2^4-1的个位数相同,为5
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128+1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128+1)
=(2^8-1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128+1)
=(2^16-1)(2^16+1)(2^32+1)(2^64+1)(2^128+1)
=(2^32-1)(2^32+1)(2^64+1)(2^128+1)
=(2^64-1)(2^64+1)(2^128+1)
=(2^128-1)(2^128+1)
=2^256 - 1
个位数 与2^4-1的个位数相同,为5
已知(4m+1)2+|2n-m-34
已知f(x)=(m^2+m)x^(m^2-2m-1),当m取何值时,
已知f(x)=(m^2+m)x^(m^2-2m-1),当m取什么值时
已知m属于R,复数z=m(m-2)/m-1+(m平方+2m-3)i,当m为何值时
已知m²+8m-1=0,求代数式(1-m)(1+m)+(m-2)²-2(m+3)²的值
已知:2m-5n=0 求下式的值 (1+n/m-m/m-n)/(1+n/m-m/m-n)
已知m-(1/m)=3.则4-m平方/2+3m/2的值为?
已知二次函数y=-x²+2(m-1)x+2m-m²
已知sinx+cosx=m,(|m|≤2,且|m|≠1),
已知m平方-5m+1=0,求(1)m三次方+1/m三次方的值 (2)m-1/m的值
已知m的绝对值为2,则(1-2m+m²)等于多少?
已知m的平方+m-1=0,则m的立方+2(m的平方)+2004=多少?