f.2數唔該 急! (20)

2009-04-13 7:45 pm
Factorize 6(4q-5r)²-10r(5r-4q)
Factorize 8aq-10cr-10ar-16bq+20br+8cq
更新1:

8aq-10cr-10ar-16bq+20br+8cq (8aq-16bq+8cq) - (10ar-20br+10cr) 呢條可以隨意組成兩group?

回答 (1)

2009-04-13 7:54 pm
✔ 最佳答案
Factorize 6(4q-5r)-10r(5r-4q)
= 6(4q-5r) + 10r(4q-5r)
= (4q-5r)[6(4q-5r) + 10r)
= (4q-5r)(24q-20r)
= 4(4q-5r)(6q-5r)
Factorize 8aq-10cr-10ar-16bq+20br+8cq
= (8aq-16bq+8cq) - (10ar-20br+10cr)
= 8q(a-2b+c) - 10r(a-2b+c)
= (8q-10r)(a-2b+c)
=2(4q-5r)(a-2b+c)


2009-04-13 11:57:50 補充:
The 1 and 2 lines missing the square (^2)

Factorize 6(4q-5r)^2 -10r(5r-4q)

= 6(4q-5r)^2 + 10r(4q-5r)

2009-04-13 23:50:00 補充:
不是隨意的,注意數字特徵: 8q : - 16q : 8q = 10r : - 20r : 10r
然後兩項都可抽(a-2b+c)


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