數x^2-8x-3=Ax(x-5)+B(x+3)+6.. 急

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

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

　　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