3x^2-6x+5=3(x-a)^2+b
這是恆等式
求a,b 的值!!
數學(恆等式)一問(!急!)
2007-08-01 5:05 am
回答 (2)
2007-08-01 5:11 am
✔ 最佳答案
3x^2-6x+5=3(x-a)^2+bRHS = 3*(x^2 - 2ax + a^2) + b
= 3x^2 - (6a)x + (3a^2 + b)
-6a = -6
a = 1
5 = (3a^2+b)
5 = 3+b
b = 2
therefore, a = 1 & b = 2
2007-08-01 5:11 am
L.H.S =3x^2 - 6x +5
R.H.S = 3 (x^2 -2ax + a^2) +b
= 3x^2 -6ax + a^2+ b
因為L.H.S =R.H.S
-6ax=-6x --------1
a^2 + b = 5 --------------2
1)
a=1 -----3
put3 into 2
1^2+ b =5
b=4
R.H.S = 3 (x^2 -2ax + a^2) +b
= 3x^2 -6ax + a^2+ b
因為L.H.S =R.H.S
-6ax=-6x --------1
a^2 + b = 5 --------------2
1)
a=1 -----3
put3 into 2
1^2+ b =5
b=4
參考: 自己
收錄日期: 2021-04-23 22:08:11
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070731000051KK05168