江湖救急!!!!一條因式分解的數學問題!!!!!20分 (如能解答,萬分感激)
因式分解下列數式
-m3n(rs-t)2 -mn2 (t-rs)3
help~~
江湖救急!!!!一條因式分解的數學問題!!!!!20分 (如能解答,萬分感激)
2006-12-06 7:29 am
回答 (3)
2006-12-06 7:50 am
✔ 最佳答案
-m^3n(rs-t)^2 -mn^2 (t-rs)^3= -m^3n(t-rs)^2 -mn^2 (t-rs)^3
= -mn (t -rs)^2 [m^2 + n(t - rs)]
= -mn [m^2 + n(t - rs)] (t -rs)^2
參考: me
2006-12-06 9:34 am
-m^3n(rs-t)^2 -mn^2 (t-rs)^3
= -m^3n(t-rs)^2 -mn^2 (t-rs)^3
= -mn (t -rs)^2 [m^2 + n(t - rs)]
= -mn (t-rs)^2 [m^2+nt-nrs]
= mn (t-rs)^2 (nrs-nt-m^2)
if -m3n(rs-t)2 -mn2 (t-rs)3
=6mn[-(rs-t)-(t-rs)]
=6mn(-rs+t-t+rs)
=6mn(0)
=0
= -m^3n(t-rs)^2 -mn^2 (t-rs)^3
= -mn (t -rs)^2 [m^2 + n(t - rs)]
= -mn (t-rs)^2 [m^2+nt-nrs]
= mn (t-rs)^2 (nrs-nt-m^2)
if -m3n(rs-t)2 -mn2 (t-rs)3
=6mn[-(rs-t)-(t-rs)]
=6mn(-rs+t-t+rs)
=6mn(0)
=0
參考: me
2006-12-06 7:35 am
係咪-m3(次)n(rs-t)2 -mn2(次)(t-rs)3 呀 ? @@
收錄日期: 2021-04-12 22:51:15
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061205000051KK05305