Maths about Factorization

2013-10-07 1:37 am
1. 243(m-n)² - 27n²

2. (2m-n)² - (m+n)²

3. (p-q)³ + (2p-q)³

4. a³ b^6 + 125 c^12 d³

回答 (1)

2013-10-07 1:54 am
✔ 最佳答案
1. 243(m-n)² - 27n²
= 27[ 9(m-n)² - n² ]
= 27[ [3(m-n)]² - n² ]
= 27[3(m-n) + n][3(m-n) - n]
= 27(3m-3n+n)(3m-3n-n)
= 27(3m-2n)(3m-4n)


2. (2m-n)² - (m+n)²
= [(2m-n) + (m+n)][(2m-n) - (m+n)]
= (2m-n + m+n)(2m-n - m-n)
= (3m)(m-2n)
= 3m(m-2n)


3. (p-q)³ + (2p-q)³
= [(p-q) + (2p-q)][(p-q)² - (p-q)(2p-q) + (2p-q)²]
= (p-q + 2p-q)[p² - 2pq +q² - (2p² - 3pq + q²) + 4p² - 4pq + q²]
= (3p-2q)(p² - 2pq +q² - 2p² + 3pq - q² + 4p² - 4pq + q²)
= (3p-2q)(3p² - 3pq + q²)


4. a³ b⁶ + 125 c^12 d³
= (ab²)³ + (5c⁴d)³
= [(ab²) + (5c⁴d)][(ab²)² - (ab²)(5c⁴d) + (5c⁴d)²]
= (ab² + 5c⁴d)(a²b⁴ - 5ab²c⁴d + 25c⁸d²)


收錄日期: 2021-04-13 19:45:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20131006000051KK00200

檢視 Wayback Machine 備份