x^2+x-k(k+1)=0
why會變成
(x-k)(x+k+1)=0
F.4 Maths
2011-12-29 10:43 pm
回答 (4)
2011-12-29 10:54 pm
✔ 最佳答案
_____x²+x-k(k+1)=0_______x²+x-k²-k=0_______x²-k²+x-k=0(x-k)(x+k)+(x-k)=0 ____(x-k)(x+k+1)=0*** Remember that a²-b²=(a-b)(a+b).
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2011-12-30 1:58 am
x^2+x-k(k+1)=0
x^2+x-k^2-k=0
(x^2-k^2)+(x-k)=0
(x+k)(x-k)+(x-k)=0
(x-k)(x+k+1)=0
x^2+x-k^2-k=0
(x^2-k^2)+(x-k)=0
(x+k)(x-k)+(x-k)=0
(x-k)(x+k+1)=0
2011-12-29 10:58 pm
用Factorization便會變成這答案:
x^2+x-k(k+1)=0
x^2+x-k^2-k=0
x^2-k^2+(x-k)=0
(x-k)(x+k)+(x-k)=0
(x-k)(x+k+1)=0
Therefore,
x^2+x-k(k+1)=0
會變成
(x-k)(x+k+1)=0
x^2+x-k(k+1)=0
x^2+x-k^2-k=0
x^2-k^2+(x-k)=0
(x-k)(x+k)+(x-k)=0
(x-k)(x+k+1)=0
Therefore,
x^2+x-k(k+1)=0
會變成
(x-k)(x+k+1)=0
2011-12-29 10:57 pm
x^2+x-k(k+1)=0
x^2+x-k^2-k=0
(x^2-k^2)+(x-k)=0
(x-k)(x+k)+(x-k)=0
(x-k)(x+k+1)=0
---------the end--------
x^2+x-k^2-k=0
(x^2-k^2)+(x-k)=0
(x-k)(x+k)+(x-k)=0
(x-k)(x+k+1)=0
---------the end--------
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111229000051KK00490