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由已知有a+1b=x,①;b+1c=x,②;c+1d=x,③;d+1a=x,④;由①解出b=1x−a⑤代入②得c=x−ax2−ax−1⑥將⑥代入③得x−ax2−ax−1+1d=x即dx3-(ad+1)x2-(2d-a)x+ad+1=0⑦由④得ad+1=ax,代入⑦得(d-a)(x…
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