第一题等式第一项乘个(2-1),在和(2+1)使用平方差公式,得(2^2-1),再和(2^2+1),使用平方差,依此类推,结果为2^64.
求解以下两式的值：1、（2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64
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相关问题
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(2+1)(2∧2+1)(2∧4+1)(2∧8+1)(2∧16+1)(2∧32+1)-2∧64求值
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(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),
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(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)=?
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计算:(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)
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(1)已知m=(2+1)*(2^2+1)*(2^4+1)*(2^8+1)*(2^16+1)*(2^32+1)*(2^64
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1/2+1/4+1/8+1/16+1/32+1/64简便运算＝﹙1／2＋1／4＋1／8＋1／16＋1／32＋1／64＋1
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(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1的末位数字.
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已知,A=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^12
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求值:(2+1)*(2^2+1)*(2^4+1)*(2^8+1)*(2^16+1)-2^32
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试确定(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)的末位数字.