(1/a+1)+(2/a^2+1)+(4/a^4+1)+(8/a^8+1)+(16/a^16+1)-(32/a^32+1)
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(1/a+1)+(2/a^2+1)+(4/a^4+1)+(8/a^8+1)+(16/a^16+1)-(32/a^32+1)
(1/a+1)+(2/a^2+1)+(4/a^4+1)+(8/a^8+1)+(16/a^16+1)-(32/a^32+1)
(1/a+1)+(2/a^2+1)+(4/a^4+1)+(8/a^8+1)+(16/a^16+1)-(32/a^32+1)
这个题上的 +1 应该是加到分母上吧 也就是这样 1/(a+1)+2/(a^2+1)+4/(a^4+1)+8/(a^8+1)+16/(a^16+1)-32/(a^32+1) = 1/(a-1)+[ -1/(a-1)+1/(a+1 ) ]+2/(a^2+1)+4/(a^4+1)+8/(a^8+1)+16/(a^16+1)-32/(a^32+1)] = 1/(a-1)+[-2/(a^2-1)+1/(a^2+1))+4/(a^4+1)+8/(a^8+1)+16/(a^16+1)-32/(a^32+1)]=1/(a-1)-32/(a^32-1) - 32/(a^32+1)=1/(a-1)+64a^32/(a^64-1)
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【1】a+a=a×a a= [ ]【2】a×a=a÷a a=[ ]【3】a×a=a-a a=[ ] [4]a-a=a+a a=[ ]
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(1-16a×a×a×a)÷(4a×a+1)÷(2a+1)
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