a ) Factorize m^4 - 2m^2n^2 + n^4 by using identity
b ) Hence , if m/n + n/m = 2 , find the value of m^4 - 2m^2n^2 + n^4
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math factorization question
2011-11-25 5:14 am
回答 (3)
2011-11-25 5:47 am
✔ 最佳答案
a) m^4 - 2m^2n^2 + n^4= (m^2)^2 - 2(m^2)(n^2) + (n^2)^2
= (m^2 - n^2)^2
= (m+n)^2 (m-n)^2
b) m/n + n/m = 2
(m^2 + n^2)/mn = 2
m^2 + n^2 = 2mn
m^2 - 2mn + n^2 = 0
(m-n)^2 = 0
m^4 - 2m^2n^2 + n^4
= (m+n)^2 (m-n)^2
= (m+n)^2 * 0
= 0
參考: Knowledge is power.
2011-11-26 12:01 am
a)
m^4 - 2(m^2)(n^2) + n^4
= (m^2 - n^2)^2
= [(m-n)(m+n)]^2
= (m-n)^2 (m+n)^2
b)
(m/n) + (n/m) = 2
(m^2) + (n^2) = 2mn
(m^2) -2mn + (n^2) = 0
(m-n)^2 = 0
so
m^4 - 2(m^2)(n^2) + n^4
= (m-n)^2 (m+n)^2
=0
m^4 - 2(m^2)(n^2) + n^4
= (m^2 - n^2)^2
= [(m-n)(m+n)]^2
= (m-n)^2 (m+n)^2
b)
(m/n) + (n/m) = 2
(m^2) + (n^2) = 2mn
(m^2) -2mn + (n^2) = 0
(m-n)^2 = 0
so
m^4 - 2(m^2)(n^2) + n^4
= (m-n)^2 (m+n)^2
=0
2011-11-25 6:53 am
m^4 - 2(m^2)(n^2) + n^4
= (m^2 - n^2)^2
= [(m-n)(m+n)]^2
= (m-n)^2 (m+n)^2
(m/n) + (n/m) = 2
(m^2) + (n^2) = 2mn
(m^2) -2mn + (n^2) = 0
(m-n)^2 = 0
thus, m^4 - 2(m^2)(n^2) + n^4 = (m-n)^2 (m+n)^2 = 0
= (m^2 - n^2)^2
= [(m-n)(m+n)]^2
= (m-n)^2 (m+n)^2
(m/n) + (n/m) = 2
(m^2) + (n^2) = 2mn
(m^2) -2mn + (n^2) = 0
(m-n)^2 = 0
thus, m^4 - 2(m^2)(n^2) + n^4 = (m-n)^2 (m+n)^2 = 0
收錄日期: 2021-04-25 17:08:50
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https://hk.answers.yahoo.com/question/index?qid=20111124000051KK00845