問題;
[(m+n)(m-n)]²
請問點計?
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更新1:
我中2的;.
數學1問!吾該晒!_!請進.
回答 (7)
2008-10-28 7:01 am
✔ 最佳答案
e 題.. 我會先將入面解開黎計解完就會變簡單
[(m+n)(m-n)]²
= (m²-n²)²
=m^4 - 2m²n² + n^4
^_^
參考: no
2008-11-02 10:54 am
似乎大家都有心fake 發問者 .... 點解會大家都有出錯 ?
2008-10-29 7:15 am
[(m+n)(m-n)]²
=[m(m-n)+n(m-n)]²
=(m²-mn+mn-n)²
=(m²-n)²
=(m²-n)(m²-n)
=m²(m²-n)-n(m²-n)
=2m²-m²n-m²n+n²
=2m²-2m²n+n²
=[m(m-n)+n(m-n)]²
=(m²-mn+mn-n)²
=(m²-n)²
=(m²-n)(m²-n)
=m²(m²-n)-n(m²-n)
=2m²-m²n-m²n+n²
=2m²-2m²n+n²
參考: me
2008-10-29 5:13 am
[(m+n)(m-n)]²
Solution: [(m+n)(m-n)]²
= [(m^2 - n^2 )]^2
= (m^2)^2 - 2 ( m^2 ) (n^2) + (n^2)^2
= m^4 - 2 (m^2)(n^2) + n^4 ( Answer )
Solution: [(m+n)(m-n)]²
= [(m^2 - n^2 )]^2
= (m^2)^2 - 2 ( m^2 ) (n^2) + (n^2)^2
= m^4 - 2 (m^2)(n^2) + n^4 ( Answer )
參考: 希望本人的解答能夠幫助到各下解決功課的疑難, 本人是”全職公文式數學導師”, 請投本人一票, 謝謝!!
2008-10-29 4:48 am
應該係展開啦@@
[(m+n)(m-n)]²
=(m²-n²)²............用(a-b)(a+b)=a²-b²
=(m²)²-2m²n²+(n²)²...........用(a-b)²=a²-2ab+b²
=m^4-2m²n²+n^4
完成~
注 : 見到有2次方就吳明既 用a=m²,b=n² 就ok了@@
[(m+n)(m-n)]²
=(m²-n²)²............用(a-b)(a+b)=a²-b²
=(m²)²-2m²n²+(n²)²...........用(a-b)²=a²-2ab+b²
=m^4-2m²n²+n^4
完成~
注 : 見到有2次方就吳明既 用a=m²,b=n² 就ok了@@
參考: 會考數學拿c既人~
2008-10-28 9:21 am
先乘括號,再自乘
[(m+n)(m-n)]²
=[(m-n)m-(m+n)n]²
=[(mm-mn-mn-nn]²
=[(mm-mn-mn-nn]²
=(mm)²-(mn)²-(mn)²-(nn)²
= (m²-n²)²
=m*4 - 2m²n² + n*4
[(m+n)(m-n)]²
=[(m-n)m-(m+n)n]²
=[(mm-mn-mn-nn]²
=[(mm-mn-mn-nn]²
=(mm)²-(mn)²-(mn)²-(nn)²
= (m²-n²)²
=m*4 - 2m²n² + n*4
參考: 唔係好識打數學符號,不過應該差不多
2008-10-28 7:29 am
[(m+n)(m-n)]²
= (m²-n²)²
=m^4 - 2m²n² + n^4
= (m²-n²)²
=m^4 - 2m²n² + n^4
收錄日期: 2021-04-29 16:13:03
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081027000051KK02315