[F.3數]Factorization(40分有答案求步驟)

2010-09-09 1:25 am
Factorize the following expressions.

(1) 128r^3+52s^3

(2) 24u^3-375v^3

(3) 2h^3-16k^3+3h-6k

(4) m^3+m^2-n^3-n^2

(5) t^6-t^3+t-1

(6) u^3+8-(u+2)^2

公式:
a^3-b^3=(a-b)(a^2+ab+b^2)
a^3+b^3=(a+b)(a^2-ab+b^2)

答案:
(1) 2(4r+3s)(16r^2-12rs+9s^2)

(2) 3(2u-5v)(4u^2+10uv+25v^2)

(3) (h-2k)(2h^2+4hk+8k^2+3)

(4) (m-n)(m^2+mn+n^2+m+n)

(5) (t-1)(t^5+t^4+t^3+1)

(6) (u+2)(u-1)(u-2)

謝謝各位幫忙。
更新1:

就係唔識做先要睇多D例子+做法, 睇下睇下就識做。

回答 (3)

2010-09-09 1:47 am
✔ 最佳答案
(1) 128r^3+54s^3= 2(64r^3 + 27s^3)= 2[(4r)^3 + (3s)^3]= 2 (4r + 3s) [(4r^2) + (4r)(3r) + (3s)^2]= 2 (4r + 3s) (16r^2 - 12rs + 9s^2) (2) 24u^3-375v^3= 3 (8u^3 - 125v^3)= 3 [(2u)^3 - (5v)^3]= 3 (2u - 5v) (4u^2 + 10uv + 25v^2) (3) 2h^3-16k^3+3h-6k= 2(h^3 - 8k^3) + 3(h - 2k)= 2(h - 2k)(h^2 + 2hk + 4k^2) + 3(h - 2k)= (h - 2k) [ 2(h^2 + 2hk + 4k^2) + 3 ]= (h - 2k)(2h^2 + 4hk + 8k^2 + 3) (4) m^3+m^2-n^3-n^2= m^3 - n^3 + m^2 - n^2= (m - n)(m^2 + mn + n^2) + (m - n)(m + n)= (m - n)(m^2 + mn + n^2 + m + n) (5) t^6-t^3+t-1= (t^3)(t^3 - 1) + t - 1= (t^3) (t - 1)(t^2 + t + 1) + (t - 1)= (t - 1) [(t^3)(t^2 + t + 1) + 1]= (t - 1)(t^5 + t^4 + t^3 + 1) (6) u^3+8-(u+2)^2= (u^3 + 2^3) - (u + 2)^2= (u + 2)(u^2 - 2u + 4) - (u + 2)^2= (u + 2) [u^2 - 2u + 4 - (u + 2)]= (u + 2) (u^2 - 3u + 2)= (u + 2) (u - 1) (u - 2)
2010-09-09 5:12 am
你講到你咁勁又唔答?
2010-09-09 1:44 am
都係應用公式係個題目到.....
有乜好問....

及咁都要問..
祝你公開考試肥佬


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