請高人指點:-
(a) 展開(4y2 次方-5y+3)(ay-1)
(b)考慮()的結果.
(i) 寫出該多項式的次數.
(ii) 若y3次方的係數為16, 寸求a的值.
多項式很難做, 請幫手
2011-08-12 2:10 am
回答 (3)
2011-08-12 7:20 am
✔ 最佳答案
(a) (4y^2-5y+3)(ay-1)= (4ay^3-5ay^2+3ay)-(4y^2-5y+3)
= 4ay^3 - (5a+4)y^2 + (3a+5)y - 3
(b)(i) 次數 = 3
(ii) 4a = 16
a = 4
2011-08-12 9:58 pm
a)(4y^2-5y+3)(ay-1)
=ay(4y^2-5y+3)-1(4y^2-5y+3)
=4ay^3-5ay^2+3ay-4y^2+5y-3
=4ay^3-(5a+4)y^2+(3a+5)y-3
(b)(i)該多項式的次數=3.
(b)(ii)由(a),y^3的係數=4a
therefore 4a=16
a=4
請 選 我 為 最 佳 ！
=ay(4y^2-5y+3)-1(4y^2-5y+3)
=4ay^3-5ay^2+3ay-4y^2+5y-3
=4ay^3-(5a+4)y^2+(3a+5)y-3
(b)(i)該多項式的次數=3.
(b)(ii)由(a),y^3的係數=4a
therefore 4a=16
a=4
請 選 我 為 最 佳 ！
2011-08-12 5:56 pm
(a)(4y^2-5y+3)(ay-1)
=ay(4y^2-5y+3)-1(4y^2-5y+3)
=4ay^3-5ay^2+3ay-4y^2+5y-3
=4ay^3-(5a+4)y^2+(3a+5)y-3
(b)(i)該多項式的次數=3.
(b)(ii)由(a),y^3的係數=4a
therefore 4a=16
a=4
=ay(4y^2-5y+3)-1(4y^2-5y+3)
=4ay^3-5ay^2+3ay-4y^2+5y-3
=4ay^3-(5a+4)y^2+(3a+5)y-3
(b)(i)該多項式的次數=3.
(b)(ii)由(a),y^3的係數=4a
therefore 4a=16
a=4
參考: just my answer
收錄日期: 2021-04-13 18:09:28
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