Simipify (4b^2+4b^3-5b+7)-(2+4b-b^3+2b^2)
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[[急急急,20分ar]]F.2 mathz 1題
2007-01-19 4:30 am
回答 (4)
2007-01-19 4:37 am
✔ 最佳答案
(4b^2+4b^3-5b+7)-(2+4b-b^3+2b^2)= (4b^3+4b^2-5b+7)-(-b^3+2b^2+4b+2)
= (4-(-1))b^3+(4-2)b^2+(-5-4)b+(7-2)
= 5b^3+2b^2-9b+5
排好次序就可以加減，好似計多位數一樣，只是加數無進位。
2007-01-19 4:59 am
(4b^2+4b^3-5b+7)-(2+4b-b^3+2b^2)
= 4b^2 + 4b^3 - 5b + 7 - 2 - 4b - b^3 + 2b^2
= 6b^2 + 3b^3 - 9b + 5
= 3b^3 + 6b^2 - 9b + 5
= 4b^2 + 4b^3 - 5b + 7 - 2 - 4b - b^3 + 2b^2
= 6b^2 + 3b^3 - 9b + 5
= 3b^3 + 6b^2 - 9b + 5
參考: me
2007-01-19 4:39 am
(4b^2+4b^3-5b+7)-(2+4b-b^3+2b^2)
(4b^2+4b^3-5b+7)-2-4b+b^3-2b^2
5b^3+2b^2-9b+5
(4b^2+4b^3-5b+7)-2-4b+b^3-2b^2
5b^3+2b^2-9b+5
2007-01-19 4:37 am
(4b² + 4b³ + 5b + 7) - (2 + 4b - b³ + 2b²)
= 4b² + 4b³ + 5b - 7 - 2 - 4b + b³ - 2b²
= (4b³ + b³) + (4b² - 2b²) + (5b - 4b) + (-7 - 2)
= 5b³ + 2b² + b - 9
2007-01-18 20:40:18 補充：
抄錯題目：(4b² + 4b³ - 5b + 7) - (2 + 4b - b³ + 2b²)= 4b² + 4b³ - 5b + 7 - 2 - 4b + b³ - 2b²= (4b³ + b³) + (4b² - 2b²) + (-5b - 4b) + (7 - 2)= 5b³ + 2b² - 9b + 5
= 4b² + 4b³ + 5b - 7 - 2 - 4b + b³ - 2b²
= (4b³ + b³) + (4b² - 2b²) + (5b - 4b) + (-7 - 2)
= 5b³ + 2b² + b - 9
2007-01-18 20:40:18 補充：
抄錯題目：(4b² + 4b³ - 5b + 7) - (2 + 4b - b³ + 2b²)= 4b² + 4b³ - 5b + 7 - 2 - 4b + b³ - 2b²= (4b³ + b³) + (4b² - 2b²) + (-5b - 4b) + (7 - 2)= 5b³ + 2b² - 9b + 5
收錄日期: 2021-04-23 19:49:40
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