→ ken1325 :非常清楚，感謝 07/19 13:19
※ 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)
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