1 X2Yn-1 + 1Yn+1 - 1 XYn
64............9.........12
[n]->次方 [x y]->未g數 [...]->冇野
Factorization 一題~~~
2009-12-09 3:01 am
回答 (2)
2009-12-09 3:49 am
If i understand your question...
[(x^2){y*(n-1)}]/64+ {y*(n+1)}/9 - (xy*n)/12
=[(x^2){y*(n-1)}]/(2^6)+{y^(n+1)}/(3^2)-(xy^n)/(2^2x3)
=[9(x^2){y^(n-1)}+64y^(n+1)-48xy^n]/(3^2x2^6)
={y^(n-1)(9x^2-48xy+64y^2)}/576
={y^(n-1)}{(3x-8y)^2}/576
={y^n(3x-8)^2}/576y //
[(x^2){y*(n-1)}]/64+ {y*(n+1)}/9 - (xy*n)/12
=[(x^2){y*(n-1)}]/(2^6)+{y^(n+1)}/(3^2)-(xy^n)/(2^2x3)
=[9(x^2){y^(n-1)}+64y^(n+1)-48xy^n]/(3^2x2^6)
={y^(n-1)(9x^2-48xy+64y^2)}/576
={y^(n-1)}{(3x-8y)^2}/576
={y^n(3x-8)^2}/576y //
2009-12-09 3:10 am
(1/64)(x^2) * y^(n-1) + (1/9) y^(n+1) - (1/12)xy^n
= [y^(n-1)] * [ (x^2)/64 + (y^2)/9 - xy/12]
= [y^(n-1)] * [ (x/8)^2 + (y/3)^2 - 2(x/8)(y/3)]
= [y^(n-1)] [ (x/8 - y/3)^2 ]
= [y^(n-1)] * [ (x^2)/64 + (y^2)/9 - xy/12]
= [y^(n-1)] * [ (x/8)^2 + (y/3)^2 - 2(x/8)(y/3)]
= [y^(n-1)] [ (x/8 - y/3)^2 ]
收錄日期: 2021-04-21 22:05:40
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091208000051KK01184