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用带余除法计算 (2x的2次方+5x-1)/(x+2)
人气:408 ℃ 时间:2020-08-29 01:08:26
解答
(2x^2+5x-1)/(x+2)
=[(2x^2+8x+8)-3x-9]/(x+2)
=[2(x^2+4x+4)-3x-9]/(x+2)
=[2(x+2)^2-(3x+9)]/(x+2)
=2(x+2)^2/(x+2)-(3x+9)/(x+2)
=2(x+2)-(3x+9)/(x+2)
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