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(2+i)x²-(5+i)x+(2-2i)=0 x是實數 2x²+x²i-5x-xi+2-2i=0 (2x²-5x+2)+(x²-x-2)i=0 則2x²-5x+2=0,x²-x-2=0同時成立 (2x-1)(x-2)=0 (x-2)(x+1)=0 公共根是x=2 所以實數解是x=2
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