(y+3)次方+3(y-1)+2
16t次方-(4t-3)(4t+3)
(a-2b)(a次方+ab+b次方)
thx!!
多項式幾條!
2006-11-29 4:37 am
回答 (2)
2006-11-29 5:05 am
✔ 最佳答案
1. (y+3)^2+3(y-1)+2= (y^2+6y+9)+(3y-3)+2
=y^2+9y+8
2. 16t^2-(4t-3)(4t+3)
= 16t^2-(16t^2+12t-12t-9)
= - 9
3. (a-2b)(a^2+ab+b^2)
= a^3+a^2b+ab^2-2a^2b-2ab^2-2b^3
= a^3-a^2b-ab^2-2b^3
2006-11-29 5:21 am
相信您所寫的次方是 2次方的意思:
1. (y+3)² + 3 (y-1) + 2
= (y² + 6y + 9) + (3y - 3) + 2
= y² + 9y + 8
= (y+8) (y+1)
2. 16t² - (4t-3) (4t+3)
= 16t² - (16t² - 9)
= 9
3. (a-2b) (a²+ab+b²)
= a³ + a²b + ab² - 2a²b - 2ab² - 2b³
= a³ - a²b - ab² - 2b³
2006-11-29 09:21:26 補充：
樓上的朋友，第2題 /- 符號攪錯了!
1. (y+3)² + 3 (y-1) + 2
= (y² + 6y + 9) + (3y - 3) + 2
= y² + 9y + 8
= (y+8) (y+1)
2. 16t² - (4t-3) (4t+3)
= 16t² - (16t² - 9)
= 9
3. (a-2b) (a²+ab+b²)
= a³ + a²b + ab² - 2a²b - 2ab² - 2b³
= a³ - a²b - ab² - 2b³
2006-11-29 09:21:26 補充：
樓上的朋友，第2題 /- 符號攪錯了!
收錄日期: 2021-04-18 20:22:01
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