唔明呢條數點計?
回答 (4)
2007-07-30 9:12 pm
✔ 最佳答案
如果是factorize [a+(b+c)][a-(b+c)]
=[a^2-(b+c)^2]
2007-07-30 23:33:32 補充：
如果是expand(a b c)(a)-(a b c)(b)-(a b c)(c)=a^2-b^2-c^2-ab ac-2bc
2007-07-30 9:16 pm
(a+b+c)(a-b-c)
= a^2 - ab - ac + ab - b^2 - bc + ac - bc - c^2
= a^2 - b^2 - c^2 - ab + ab - ac + ac - bc - bc
= a^2 - b^2 - c^2 - 2bc
= a^2 - ab - ac + ab - b^2 - bc + ac - bc - c^2
= a^2 - b^2 - c^2 - ab + ab - ac + ac - bc - bc
= a^2 - b^2 - c^2 - 2bc
2007-07-30 9:14 pm
a2-ab-ac+ab-b2-bc+ac-bc-c2
=a2-b2-c2-bc
=a2-b2-c2-bc
2007-07-30 9:13 pm
(a+b+c)(a-b-c)
= (a+b+c)(a)-(a+b+c)(b)-(a+b+c)(c)
= aa + ab + ac - ab - bb - bc - ab - cb - cc
= aa - ab + ac - bb - 2bc - cc
= (a+b+c)(a)-(a+b+c)(b)-(a+b+c)(c)
= aa + ab + ac - ab - bb - bc - ab - cb - cc
= aa - ab + ac - bb - 2bc - cc
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