請幫我解決以下問題，
1) 因式分解(x+1)(x-1) - (x+1)(2x+1)
2) 已知二次方程 x^2+(k+1)x+9=0 有兩等根，求常數 k 的值。
3) 求 p 的值使二次方程 2x^2+2p=(4p+1)x 有兩相同實根。
thanks^_^
maths唔識ar~~
2006-12-11 1:15 am
回答 (2)
2006-12-11 1:34 am
✔ 最佳答案
1) (x+1)(x-1) - (x+1)(2x+1)=(x^2-1)-(2x^2+x+2x+1)
=x^2-1-2x^2-x-2x-1
=x^2-2x^2-x-2x-1-1
= -x^2-3x-2
= -(x^2+3x+2) [呢個步驟做唔做都會岩]
2) 因為有兩等根,
判別式=0
(k+1)^2 - 4(1)(9)=0
k^2+2k+1-36=0
k^2+2k-35=0
(k-5)(k+7)=0
k-5=0 或 k+7=0
k=5 或 k= -7
3) 2x^2+2p=(4p+1)x
2x^2+2p=4px+x
2x^2+2p-4px-x=0
2x^2-(4p+1)x+2p=0 [抽x前既項出黎要小心正負號]
因為有兩相同實根,
判別式=0
[-(4p+1)]^2-4(2)(2p)=0
(4p+1)^2-16p=0
16p^2+8p+1-16p=0
16p^2-8p+1=0
(4p-1)^2=0
4p-1=0
4p=1
p=1/4
參考: me
2006-12-11 1:26 am
1)
[ ( X + 1 ) X - ( X + 1 ) 1 ] - [ ( X + 1 ) 2X + ( X + 1 )1]
=( X^2 + X - X - 1 ) - ( 2X^2 + 2X + X+1)
=( X^2 - 1 ) - ( 2X^2 + 3X + 1)
=X^2 - 1 -2X^2 + 3X +1
=X^2 - 2X^2 +3X -1 +1
= -X^2 +3X
註:例 x^2即是x既2次方
[ ( X + 1 ) X - ( X + 1 ) 1 ] - [ ( X + 1 ) 2X + ( X + 1 )1]
=( X^2 + X - X - 1 ) - ( 2X^2 + 2X + X+1)
=( X^2 - 1 ) - ( 2X^2 + 3X + 1)
=X^2 - 1 -2X^2 + 3X +1
=X^2 - 2X^2 +3X -1 +1
= -X^2 +3X
註:例 x^2即是x既2次方
參考: 自己=_=v
收錄日期: 2021-04-12 21:02:03
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061210000051KK03424