解下列各一元二次方程式:
12分之1(x-3)^2=3分之1x
(x+2)^2=(2x-1)(x+2)
[請使用因式分解法解一元二次方程式]
不會算十字交乘法
2012-12-21 4:45 am
回答 (2)
2012-12-21 4:52 am
✔ 最佳答案
解下列各一元二次方程式:12分之1(x-3)^2=3分之1x
==>(x-3)^2/12 -(4x/12)=0
==>[(x-3)^2-4x]/12=0
==>x^2-10x+9=0
==>(x-9)(x-1)=0
==>x=9,x=1
(x+2)^2=(2x-1)(x+2)
[請使用因式分解法解一元二次方程式]
==>(x+2)^2-(2x-1)(x+2)=0
==>(x+2)[(x+2)-(2x-1)]=0
==>(x+2)(-x+3)=0
==>-(x+2)(x-3)=0
==>x=-2,x=3
2012-12-21 5:42 am
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參考: 自己
收錄日期: 2021-05-03 11:51:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20121220000015KK07606