(2^2 +1)(2^4 +1)(2^8 +1)...(2^32 +1) 已知x^2-y^2=8,x+y=4,求x,y的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/29 10:50:02
x)03R6&PD`97="H2BBDƦ
Z6I*ҧYv62J[C}c-c-4Uj
rj .3]ؿ
044&b+mkaQq {5_,_t
[cJ[C<;PD dc
(2^2 +1)(2^4 +1)(2^8 +1)...(2^32 +1) 已知x^2-y^2=8,x+y=4,求x,y的值
(2^2 +1)(2^4 +1)(2^8 +1)...(2^32 +1) 已知x^2-y^2=8,x+y=4,求x,y的值
(2^2 +1)(2^4 +1)(2^8 +1)...(2^32 +1) 已知x^2-y^2=8,x+y=4,求x,y的值
(2^2 +1)(2^4 +1)(2^8 +1)...(2^32 +1)
=1/3*3*(2^2 +1)(2^4 +1)(2^8 +1)...(2^32 +1)
=1/3(2^2 -1)*(2^2 +1)(2^4 +1)(2^8 +1)...(2^32 +1)
=1/3(2^4 -1)*(2^4 +1)(2^8 +1)...(2^32 +1)
...
=1/3(2^64-1)
x+y=4 (1)
x^2-y^2=(x+y)(x-y)=4(x-y)=8
x-y=2 (2)
联立(1)(2)解得x=3,y=1
计算:(2+1)(2^2+1)(2^4+1)(2^8+1)-2^16.计算:(2+1)(2^2+1)(2^4+1)(2^8+1)-2^16.
(2+1)({2}^{2}+1)({2}^{4}+1)({2}^{8}+1).({2}^{64}+1)+1
巧算((2^1+1)(2^2+1)(2^4+1)(2^8+1)+1)/2^15
计算(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1
计算1/(1+2)+2/(1+2^2)+4/(1+2^4)+8/(1+2^8)+16/(1+2^16)
(2^2+1)*(2^4+1)*(2^8+1)*(2^16+1)*(2^32+1)
(2+1)(2-1)(2^2+1)(2^4+1).(2^8+1)化简
(1+2)*(1+2^2)*(1+2^4)*(1+2^8)*(1+2^16)
化简(1+2)(1+2^2)(1+2^4)(1+2^8)(1+2^16)
化简(2+1)(2^2+1)(2^4+1)(2^8+1)…(2^256+1)
(2+1)(2^2+1)(2^4+1)(2^8+1)……(2^1024+1)
(2+1)(2^2+1)(2^4+1)(2^8+1).(2^2048+1)=?
初中数学题&(1+2)*(1+2)^2*(1+2)^4*(1+2)^8*(1+2)^16
计算:(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)
(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
3(2 ^2+1)(2^4+1)(2^8+1)(2^16+1)-2^32
(2^2-1)(2^2+1)(2^4+1)(2^8+1) 等于多少
(2+1)(2^2+1)(2^4+1)(2^8+1)-2^16