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※ 引述《cecilia0305 (Cecilia)》之銘言: : 1.求 (2y+1)(3y-1)(4y-1)(y+1)-36y^4的因式分解 = [(2y+1)(4y-1)][(y+1)(3y-1)]-36y^4 = (8y^2+2y-1)(3y^2+2y-1) - 36y^4 = 24y^4 + 11y^2(2y-1) + (2y-1)^2 - 36y^4 = -12y^4 + 11y^2(2y-1) + (2y-1)^2 12y^2 \/ (2y-1) -1y^2 /\ (2y-1) = (12y^2+2y-1)(-y^2+2y-1) = -(12y^2+2y-1)(y-1)^2 (前一因式由一次因式檢驗法知無整係數一次因式) : 2.求(x+y)(x+1)(y+1)+xy的因式分解 = (x+y)(xy+x+y+1)+xy = (x+y)^2 + (x+y)(xy+1) + xy (x+y) \/ xy (x+y) /\ 1 = (x+y+xy)(x+y+1) : 高雄某私立國中講義題目阿> < -- LPH [acronym] = Let Program Heal us -- New Uncyclopedian Dictionary, Minmei Publishing Co. -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 114.41.35.96
cecilia0305 :挖好迅速>< 感恩~~ 10/20 17:53
Heaviside :有神 推一個! 10/20 20:32