(0.3p-0.2)(0.09p^2+0.06p+0.04)
用恆等x展開
2006-12-25 16:41:10 · 3 個解答 · 發問者 ? 1 in 科學 ➔ 數學
(0.3p-0.2)(0.09p2+0.06p+0.04)
=(0.3p-0.2)[(0.3p)2+(0.3p)(0.2)+(0.2)2]
=(0.3p)3-(0.2)3 by using 恆等式*
=0.027p3-0.008
↑
恆等式* : x3-y3 ≡(x-y)(x2+xy+y2)
2006-12-25 20:42:52 · answer #1 · answered by peterlau621 7 · 0⤊ 0⤋
(0.3p-0.2)(0.09p^2+0.06p+0.04)
=(0.3p-0.2)[(0.3p)^2+(0.3)(0.2)p+(0.2)^2]
=(0.3p-0.2)(0.3p+0.2)^2
=[(0.3p)^2-0.2^2](0.3p+0.2)^
=(0.09p-0.04)(0.3p+0.2)
2006-12-25 23:26:46 補充：
錯左(0.3p-0.2)(0.09p^2+0.06p+0.04)=(0.3p-0.2)[(0.3p)^2+(0.3p)(0.2)+(0.2)^2]=(0.3p)^3-0.2^3=0.027p^3-0.008↑x^3-y^3=(x-y)(x^2+xy+y^2)
2006-12-25 18:20:31 · answer #2 · answered by sr 6 · 0⤊ 0⤋
0.027p^3-0.008
2006-12-25 16:46:49 · answer #3 · answered by 景昆 2 · 0⤊ 0⤋