作业帮计算：（x-1/2)(2x+1)(2x∧2+1/2)(4x∧4+1/4)(16x∧8+1/16)/(256x∧16-1/256)
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/24 05:44:16
计算：（x-1/2)(2x+1)(2x∧2+1/2)(4x∧4+1/4)(16x∧8+1/16)/(256x∧16-1/256)
计算：（x-1/2)(2x+1)(2x∧2+1/2)(4x∧4+1/4)(16x∧8+1/16)/(256x∧16-1/256)
计算：（x-1/2)(2x+1)(2x∧2+1/2)(4x∧4+1/4)(16x∧8+1/16)/(256x∧16-1/256)
原式=(2x^2-1/2)(2x^2+1/2)×( )( )/(256x^16-1/256)
=(4x^4-1/4)(4x^4+1/4)( )/(256x^16-1/256)
=(16x^8-1/16)(16x^8+1/16)/(256x^16-1/256)
=(256x^16-1/256)/(256x^16-1/256)
=1
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