Identities and Factorization:
Expand the following expressions: (2)=2次方
1)-3(2d+e)(e-2d)
2)-4(2m-5n)(2)
Maths!!!!FAST!!!!
2007-10-30 7:08 am
回答 (4)
2007-10-30 7:15 am
✔ 最佳答案
1)-3(2d+e)(e-2d)=(-6d-3e)(e-2d)
=12d^2-3e^2
2)-4(2m-5n)^2
=-4(4m^2-20mn+25n^2)
=-16m^2+80mn-100n^2
2007-10-30 7:16 am
1) -3(2d+e)(e-2d) ^2 = 2次方
=-3(e+2d)(e-2d)
=-3(e^2-4d^2)
=-3e^2 + 12d^2
2) -4(2m-5n)^2
=-4(4m^2-20mn+25n^2)
=-16m^2 + 80mn - 100n^2
=-3(e+2d)(e-2d)
=-3(e^2-4d^2)
=-3e^2 + 12d^2
2) -4(2m-5n)^2
=-4(4m^2-20mn+25n^2)
=-16m^2 + 80mn - 100n^2
參考: ME
2007-10-30 7:16 am
我用^2 代表2次方
1) -3(2d+e)(e-2d)
= -3 [2de-4d^2+e^2-2de]
= -3 [e^2-4d^2]
= 12d^2-3e^2
2) -4(2m-5n)^2
= -4 [(2m)^2-2(2m)(5n)+(5n)^2]
= -4 [4m^2-20mn+25n^2]
= 80mn-16m^2-100n^2
(提提你, (a-b)^2 = a^2-2ab+b^2)
1) -3(2d+e)(e-2d)
= -3 [2de-4d^2+e^2-2de]
= -3 [e^2-4d^2]
= 12d^2-3e^2
2) -4(2m-5n)^2
= -4 [(2m)^2-2(2m)(5n)+(5n)^2]
= -4 [4m^2-20mn+25n^2]
= 80mn-16m^2-100n^2
(提提你, (a-b)^2 = a^2-2ab+b^2)
參考: 自己
2007-10-30 7:16 am
1. -3(2d+e)(e-2d)
= 3(2d+e)(2d-e)
= 3[(2d)^2 - e^2]
=3(4d^2 - e^2)
= 12d^2 - 3e^2
2. -4(2m-5n)^2
= -4[(2m)^2 - 2(2m)(5n)+(5n)^2]
=-4(4m^2 - 20mn + 25n^2)
= -16m^2 + 80mn - 100n^2
= 3(2d+e)(2d-e)
= 3[(2d)^2 - e^2]
=3(4d^2 - e^2)
= 12d^2 - 3e^2
2. -4(2m-5n)^2
= -4[(2m)^2 - 2(2m)(5n)+(5n)^2]
=-4(4m^2 - 20mn + 25n^2)
= -16m^2 + 80mn - 100n^2
收錄日期: 2021-04-13 14:15:18
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https://hk.answers.yahoo.com/question/index?qid=20071029000051KK04688