(16e^2+1)^2 - 64e^2
答案係(4e + 1)^2(4e - 1)^2
點計,謝,TEST用 TT
急,1條F2 MATHS題(20分+只限30MIN)
2011-09-28 5:36 am
回答 (2)
2011-09-28 5:54 am
✔ 最佳答案
(16e^2+1)^2 - 64e^2= [(16e^2 + 1) - 8e] [(16e^2 + 1) + 8e] -----------> (1)
= (16e^2 + 1 - 8e) (16e^2 + 1 + 8e) ---------> expand
= (4e - 1)^2 (4e + 1)^2 -----------> use cross method
(1) : (a - b)(a + b) = a^2 - b^2
Hope I can help you.
2011-09-28 5:54 am
(16e^2+1)^2 - 64e^2
=(16e^2+1+8e)(16e^2+1-8e)
=(16e^2+8e+1)(16e^2-8e+1)
=(4e + 1)^2(4e - 1)^2
ps唔明可以問我~
=(16e^2+1+8e)(16e^2+1-8e)
=(16e^2+8e+1)(16e^2-8e+1)
=(4e + 1)^2(4e - 1)^2
ps唔明可以問我~
參考: me
收錄日期: 2021-04-23 20:45:18
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https://hk.answers.yahoo.com/question/index?qid=20110927000051KK00896