因式分解以下題目:
1. 3(su-4sv)^2-2s(4tr-tu)
2. x^6+y^9
因式分解以下題目
2009-11-29 8:18 am
回答 (2)
2009-11-29 9:43 am
✔ 最佳答案
1. 3(su-4sv)^2-2s(4tv-tu)=3s^2(u-4v)^2-2st(4v-u)
=3s^2(u-4v)^2+2st(u-4v)
=s(u-4v)[3s(u-4v)+2t]
=s(u-4v)(3su-12sv+2t)
2. x^6+y^9
=(x^2+y^3)(x^4-x^2y^3+y^6)
2009-11-29 10:41 am
1.
3(su-4sv)^2-2s(4tv-tu)
=3[s(u-4v)]^2-2s[t(4v-u)]
=3s^2(u-4v)^2+2st(u-4v)
=s(u-4v)[3s(u-4v)+2t]
2.
x^6+y^9
=(x^2)^3+(y^3)^3
Let X=x^2 and Y=y^3
=X^3+Y^3
=(X+Y)(X^2-XY+Y^2)
=(x^2+y^3[x^4-(x^2)(y^3)+y^6]
2009-11-29 02:44:35 補充：
=(x^2+y^3)[x^4-(x^2)(y^3)+y^6]
3(su-4sv)^2-2s(4tv-tu)
=3[s(u-4v)]^2-2s[t(4v-u)]
=3s^2(u-4v)^2+2st(u-4v)
=s(u-4v)[3s(u-4v)+2t]
2.
x^6+y^9
=(x^2)^3+(y^3)^3
Let X=x^2 and Y=y^3
=X^3+Y^3
=(X+Y)(X^2-XY+Y^2)
=(x^2+y^3[x^4-(x^2)(y^3)+y^6]
2009-11-29 02:44:35 補充：
=(x^2+y^3)[x^4-(x^2)(y^3)+y^6]
收錄日期: 2021-04-15 22:39:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091129000051KK00035