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计算(1+1/2;)*(1+1/2²)*(1+1/2四次方)*(1+1/2八次方)+1/2十五次方
人气:178 ℃ 时间:2020-03-23 08:38:47
解答
(1+1/2;)*(1+1/2²)*(1+1/2四次方)*(1+1/2八次方)+1/2十五次方
=2*(1-1/2)[(1+1/2;)*(1+1/2^2;)*(1+1/2^4)*(1+1/2^8)]+1/2^15
=2*(1-1/2^2)(1+1/2^2;)*(1+1/2^4)*(1+1/2^8)+1/2^15
=2*(1-1/2^4)(1+1/2^4)*(1+1/2^8)+1/2^15
=2*(1-1/2^8)(1+1/2^8)+1/2^15
=2*(1-1/2^16)+1/2^15
=2-1/2^15+1/2^15
=2
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