【急】FACTORIZATION 因式分解 20分!

2008-07-11 5:13 am
有兩題

(mn-1)^2 + (m+n-2mn)(m+n-2)-1

2x^3y - 4x^2y^3 - 3z^3 + 6y^2z^3

肯定無打錯

請幫手fact左去..

回答 (2)

2008-07-12 4:00 am
✔ 最佳答案
1.(mn-1)^2+(m+n-2mn)(m+n-2)-1
=(mn)^2-2mn+1+(m+n)^2-2(m+n)-2mn(m+n)+4mn-1
=(mn)^2+2mn+(m+n)^2-2(m+n)-2mn(m+n)
=(mn-m-n)^2+2(mn-m-n)
=(mn-m-n)(mn-m-n+2)
2.2x^3y - 4x^2y^3 - 3z^3 + 6y^2z^3
=(2x^3y - 4x^2y^3)-(3z^3 + 6y^2z^3)
=2x^2y(x-2y^2)-3z^3(1-2y^2)
沒有了...
若是2x^2y - 4x^2y^3 - 3z^3 + 6y^2z^3,就可計下去。
*2.* 2x^2y - 4x^2y^3 - 3z^3 + 6y^2z^3
=2x^2y(1-2y^2)-3z^3(1-2y^2)
=(2x^2y-3z^3)(1-2y^2)
2008-07-11 3:54 pm
1. Let mn= x, m+n=y. The expression becomes
(x-1)^2 + (y-2x)(y-2) - 1
= x^2 + 1 -2x +y^2 -2y -2xy +4x -1
= x^2 -2xy +y^2 +2x -2y
= (x-y)^2 +2(x -y)
=(x-y+2)(x-y)
= (mn -m-n +2)(mn -m-n).

2008-07-11 08:01:33 補充:
For Q.2, is the first term really 2x^3y and not 2x^2y?


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