x^2-8x-3=Ax(x-5)+B(x+3)+6
常數a,b的值
數x^2-8x-3=Ax(x-5)+B(x+3)+6.. 急
2008-11-03 6:49 am
回答 (3)
2008-11-03 7:19 am
✔ 最佳答案
x^2-8x-3=Ax(x-5)+B(x+3)+6常數a,b的值
x^2-8X-3 = Ax^2-5Ax+Bx+3B+6
= Ax^2-(5A-B)x-(-3B-6)
(1) A=1
(2) 5A-B=8
(3) -3B-6=3
(2) 5(1)-B=8
5-8=B
B=-3
(3) PROOF: -3(-3)-6 = 9-6 = 3
2008-11-03 10:21 pm
x²-8x-3恆=Ax(x-5)+B(x+3)+6.................要恆等式先有得玩@@
Ax(x-5)+B(x+3)+6
=Ax²-5Ax+Bx+3B+6..........展開
=Ax²-(5A-B)x+3B+6
x²-8x-3恆=Ax²-(5A-B)x+3B+6...........a既次方一樣,係數相同
A=1
8=5A-B
8=5(1)-B
B=-3
所以常數a,b的值為1,-3
Ax(x-5)+B(x+3)+6
=Ax²-5Ax+Bx+3B+6..........展開
=Ax²-(5A-B)x+3B+6
x²-8x-3恆=Ax²-(5A-B)x+3B+6...........a既次方一樣,係數相同
A=1
8=5A-B
8=5(1)-B
B=-3
所以常數a,b的值為1,-3
參考: 會考數學拿c既人~
2008-11-03 7:32 am
x2-8x-3=Ax(x-5)+B(x+3)+6
x2-8x-3=Ax2-5A+Bx+3B+6
x2-8x-3=Ax2+Bx+(6-5A+3B)
所以A=1,B=-8
x2-8x-3=Ax2-5A+Bx+3B+6
x2-8x-3=Ax2+Bx+(6-5A+3B)
所以A=1,B=-8
參考: 自己
收錄日期: 2021-04-23 23:12:56
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https://hk.answers.yahoo.com/question/index?qid=20081102000051KK02569