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1、降次解一元二次方程降次解一元二次方程-因式分解法因式分解法Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2019-2019 Aspose Pty Ltd.解：原方程可解：原方程可变变形形为为(x+2)(x(x+2)(x2)=02)=0X+2=0X+2=0或或 x x2=02=0 x1=-2 ,x2=2 x1=-2 ,x2=2X2X24=04=0一移一移项项二分解二分解三降次三降次四求解四求解探求新知探求新知Evaluation only.Created with
2、Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2019-2019 Aspose Pty Ltd.快速回答：以下各方程的根分别是多少？快速回答：以下各方程的根分别是多少？Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2019-2019 Aspose Pty Ltd.例例1 1、解以下方程、解以下方程1 1、x2x23x3x10=0 210=0 2、(x+3)(x(x+3)(x1)=51)=5解
3、：原方程可解：原方程可变变形形为为 解：原方程可解：原方程可变变形形为为 (x(x5)(x+2)=0 x2+2x5)(x+2)=0 x2+2x8=0 8=0 x x5=05=0或或x+2=0 (xx+2=0 (x2)(x+4)=02)(x+4)=0 x1=5 ,x2=-2 x x1=5 ,x2=-2 x2=02=0或或x+4=0x+4=0 x1=2 ,x2=-4 x1=2 ,x2=-4Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2019-2019 Aspose
4、Pty Ltd. 例例2、解以下方程、解以下方程 Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2019-2019 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2019-2019 Aspose Pty Ltd.(2)(3x+1)2(2)(3x+1)25=0 5=0 解：原方程可解：原方程可变变形形为为
5、(3x+1+(3x+1+)(3x+1)(3x+1)=0 )=0 3x+1+3x+1+=0或3x+1=0 =0 x1= x1= , x2= Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2019-2019 Aspose Pty Ltd.1.1.方程右方程右边边不不为为零的化零的化为为 。零零一次因式一次因式有一个有一个一元一次方程的解一元一次方程的解 用因式分解法解一元二次方程的步骤用因式分解法解一元二次方程的步骤: :4.4.两个两个 就是原方程就是原方程 的解。
6、的解。 3.3.至少至少 一次因式一次因式为为零，得零，得 到两个一元一次方程。到两个一元一次方程。2.2.将方程左将方程左边边分解成两个分解成两个_ _ 的乘的乘积积。课堂小结课堂小结Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2019-2019 Aspose Pty Ltd.三、稳定练习三、稳定练习课本课本P40P40练习练习1 1 课堂课堂训练训练Evaluation only.Created with Aspose.Slides for .NET 3.5
7、 Client Profile 5.2.0.0.Copyright 2019-2019 Aspose Pty Ltd.1.用因式分解法解关于 的方程四、课堂延伸四、课堂延伸Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2019-2019 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2019-2019 A
8、spose Pty Ltd.解：原方程可变形为：(xa+b)(xab)=0xa+b=0 或 xab=0x1=abx2=a+b(xa)2b2=0Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2019-2019 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2019-2019 Aspose Pty Ltd.右
9、化零左分解右化零左分解两因式各求解两因式各求解简记简记歌歌诀诀：Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2019-2019 Aspose Pty Ltd.六、作业设计六、作业设计课本课本P43P43习题习题22.222.2第第6 6题题作业作业Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2019-2019 Aspose Pty Ltd.Evaluation only.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Copyright 2019-2019 Aspose Pty Ltd.
