(急)因式分解代數問題

2009-09-27 4:30 pm
可唔可以比個計算過程我


1. 6(a+1)(2次方) + 2a(a-1)

2. 8(x+2)(x-4) – 6(x-2)(x-4)


3. 15(2x+1)(2次方 ) (3y-2)(2次方) – 10(2x+1)(3次方) (3y-2)(2次方)

回答 (1)

2009-09-27 4:45 pm
✔ 最佳答案
(1) 6(a + 1)^2 + 2a(a - 1)
= 6(a^2 + 2a + 1) + 2a^2 - 2a
= 6a^2 + 12a + 6 + 2a^2 - 2a
= 8a^2 + 10a + 6
= 2(4a^2 + 5a + 3) [不能再分解,請核對題目]
(2) 8(x + 2)(x - 4) - 6(x - 2)(x - 4)
= 2(x - 4)[4(x + 2) - 3(x - 2)]
= 2(x - 4)(4x + 8 - 3x + 6)
= 2(x - 4)(x + 14)
(3) 15(2x + 1)^2(3y - 2)^2 - 10(2x + 1)^3(3y - 2)^2
= 5(2x + 1)^2(3y - 2)^2[3 - 2(2x + 1)]
= 5(2x + 1)^2(3y - 2)^2(3 - 4x - 2)
= 5(2x + 1)^2(3y - 2)^2(1 - 4x)


收錄日期: 2021-04-23 23:18:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090927000051KK00284

檢視 Wayback Machine 備份