數學一問...
回答 (3)
2007-11-30 5:43 am
✔ 最佳答案
20p(3p-2)^2-20(3p-2)p-5p= -5p [4p(3p-2)^2+4p(3p-2)+1 ]
let (3p-2) be x
= -5p (4px^2 + 4px + 1 )
= -5p (2px+1)^2
= -5p [2p(3p-2)+1)^2]
= -5p (6p^2-4p+1)^2
2007-11-30 10:23 am
- 20 p ( 3p - 2 ) ^ 2 - 20 ( 3p - 2 ) p - 5p
= - 5 p [ 4 ( 3p - 2 ) ^ 2 + 4 ( 3p - 2 ) + 1]
= - 5p (2(3p-2) + 1)^2
= - 5p ( 6p - 4 + 1 )^2
= - 5p ( 6p - 3 ) ^2
= - 5p [ 3 ( 2p - 1)]^2
= - 5p [ 9(2p - 1)^2 ]
= - 45p ( 2p - 1 ) ^ 2
= - 5 p [ 4 ( 3p - 2 ) ^ 2 + 4 ( 3p - 2 ) + 1]
= - 5p (2(3p-2) + 1)^2
= - 5p ( 6p - 4 + 1 )^2
= - 5p ( 6p - 3 ) ^2
= - 5p [ 3 ( 2p - 1)]^2
= - 5p [ 9(2p - 1)^2 ]
= - 45p ( 2p - 1 ) ^ 2
2007-11-30 7:23 am
-20p(3p-2)^2-20(3p-2)p-5p
= -5p (4(3p-2)^2 + 4(3p-2) + 1)
= -5p (2(3p-2) + 1)^2
= -5p ( 6p - 4 + 1 )^2
= -5p ( 6p - 3 ) ^2
= -5p [3(2p-1)]^2
= -5p * 3^2 * (2p-1)^2
= -45p(2p-1)^2
= -5p (4(3p-2)^2 + 4(3p-2) + 1)
= -5p (2(3p-2) + 1)^2
= -5p ( 6p - 4 + 1 )^2
= -5p ( 6p - 3 ) ^2
= -5p [3(2p-1)]^2
= -5p * 3^2 * (2p-1)^2
= -45p(2p-1)^2
收錄日期: 2021-04-13 14:36:54
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