(2+1)×(2^2+1)×(2^4+1)×(2^8+1)×(2^16+1)×(2^32+1)=2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)=(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)=(2^8-1)(2^8+1)(2^16+1)(2^32+1)=(2^16-1)(2^16+1)(2^32+1)=(2^32-1)(2^32+1)=2^64-14x^2+kx+9是
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/29 05:52:17
(2+1)×(2^2+1)×(2^4+1)×(2^8+1)×(2^16+1)×(2^32+1)
=2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)
=(2^8-1)(2^8+1)(2^16+1)(2^32+1)
=(2^16-1)(2^16+1)(2^32+1)
=(2^32-1)(2^32+1)
=2^64-1
4x^2+kx+9是完全平方式,k=12或负12
第一题在原式前面乘以(2-1)然后再利用平方差公式
第二题因为是完全平方式 所以4x^2+kx+9=(2x±3)^2 所以k=±12
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15
(1+1/2)(1+1/2^2)(1+1/2^4)^(1+1/2^32)
(2^2+1)*(2^4+1)*(2^8+1)*(2^16+1)*(2^32+1)
计算:(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)(2^2048+1),
(2+1)(2^2-1)(2^4+1)(2^8+1)(2^16-1)(2^32+1)(2^64+1)
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15=?
(-2k-1)^2+4(-k^2+1)
(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
计算:(1+2)(1+2^2)(1+2^4)(1+2^8)(1+2^16)(1+2^32)(1+2^64)
化简(1+2^(-1/32))(1+2^(-1/16))(1+2^(-1/8))(1+2^(-1/4))(1+2^(-1/2))
-1/2+(-1/6)-(-1/4)-2/3
求和:1+(1+2)+(1+2+4)+(1+2+4+6)+(1+2+4+.2^n-1)
计算:(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)/(2^32)-1=?
lim(1-1/2n).(1+1/2+1/4+.+1/2^n)
(2+1)(2^2+1)(2^4+1)x…x(2^2n+1)
(2+1)(2^2+1)(2^4+1)……(2^2n+1)
计算:(2+1)(2*2+1)(2*4+1)……(2*2n+1).