计算:(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)/(2^32)-1=?
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:(2+1)(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)(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]
=1
=(2-1)(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]
=1
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