x(x+C) = (x-1)^2 +D
Find C and D.
Finding unknowns in identities
2008-11-15 9:56 pm
回答 (2)
2008-11-15 10:05 pm
✔ 最佳答案
x(x+C) = (x-1)^2 +DL.H.S=X^2+CX
R.H.S=x^2-2X+1+D
so C=-2 , 1+D=0
D=-1
2008-11-15 14:07:11 補充：
(x-1)^2
d度用
(a-b)^2=a^2-2ab+b^2
參考: 自己的經驗[全班第1名,99%係岩ge]
2008-11-16 7:18 am
x(x+C) = (x-1)^2 +D
L.H.S=x^2+Cx
R.H.S=x^2-2x+1+D
by comparing coefficient of x
C=-2 ,
by comparing coefficient of x^0,
1+D=0
D=-1
L.H.S=x^2+Cx
R.H.S=x^2-2x+1+D
by comparing coefficient of x
C=-2 ,
by comparing coefficient of x^0,
1+D=0
D=-1
收錄日期: 2021-04-23 20:22:50
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081115000051KK00888