Re: [中學] 因式分解

看板Math作者 (艷陽天。)時間12年前 (2013/07/19 02:47), 編輯推噓0(001)
留言1則, 1人參與, 最新討論串38/78 (看更多)
※ 引述《ken1325 (早餐店帥哥)》之銘言: : xyz-(yz+xz+xy)+(z+y+x)-1 = (x-1)(y-1)(z-1) : 請問要怎麼把左式分解成右式? : thx 單純 分組提公因式 xyz - yz - xz - xy + z + y + x - 1 = ( xyz - yz - xz + z )+( - xy + y + x - 1) (8項分組=> 4,4 => 有z,沒z) = z( xy - y - x + 1 ) - (xy - y - x + 1 ) = ( xy - y - x + 1 )(z-1) = {( xy - y ) + ( -x + 1 )}(z-1) (前括弧4項 => 2,2 => 有y,沒y) = {y( x - 1 ) - ( x - 1 )}(z-1) = (x-1)(y-1)(z-1) xyz - yz - xz - xy + z + y + x - 1 = (xyz - yz) + (-xz + z) + (-xy + y) + (x - 1) (8項分組=> 2,2,2,2 => 經驗..) = yz(x-1) - z(x-1) - y(x-1) + (x-1) = (x-1){yz - z - y + 1} = (x-1){(yz - z) + (-y + 1)} (後括弧4項 => 2,2 => 有z,沒z) = (x-1){z(y - 1) - (y - 1)} = (x-1)(y-1)(z-1) -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 112.104.70.213

07/19 13:19, , 1F
非常清楚,感謝
07/19 13:19, 1F
文章代碼(AID): #1Hw3WzI2 (Math)
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文章代碼(AID): #1Hw3WzI2 (Math)