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1、ax2 + bx + c 0ax2 + bx + c 0(a0)xyox1 x2xyox1 x2xyox1 x2xyox1 x2xyox1 x2xyox1 x2Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspos
2、e Pty Ltd.解一元一次不等式时，你们知道应 具备什么知识吗？1) 若 ad ，则acd c2) 若 ad 且c 0，则a cd c3) 若 ad 且c 0，则a cd cEvaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2
3、011 Aspose Pty Ltd.还有一种数学方法可以解不等式， 你们知道吗？数形结合在解不等式中起着非常优 越的作用！先看它在解一元一次不 等式的应用吧！xyo3.5和 解方程 2x-7=0 作函数 y=2x-7 的图象 解不等式: 2x-702x-70数形结合数形结合(y0)(y0)Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2
4、.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.你们现在可以用数形结合解一元二 次不等式了吗？ 解方程 x2 x 6=0 作函数 y= x2 x 6 的图象 解不等式: -23-6xyox2 x 60x2 x 60Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Clien
5、t Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.方程： ax2+bx+c=0 的解情况函数： y=ax2+bx+c 的图图象不等式的解集 ax2+bx+c0ax2+bx+c0a0xyox1x2xo x0yxoy当0 时 ，方程有两 不等的根：x1 ，x2当0 时 ，方程有一 根 ： x0当0 时 ，方程无解xxx1 或 xx2 xxx0Rxx1xx2 Evaluation only.Evaluation only. Created with Aspose.Slide
6、s for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.1) 2x2 3x 202) 3x2 6x 23) 4x2 4x 104) x2 2x 30Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Pr
7、ofile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.2) 数轴上标根后，必须遵循 的顺序填入+、-号1) 每个因式中的 x 的系数必须是0.52+-(2x+1)(x2)0可解高次不等式1、 函数图象法 2、 数轴标根法 Evaluation only.Evaluation only. Created with Aspose.Slides for .NET
8、 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.Rx2或x-1-1x2-12-2-2a61Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.S
9、lides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose
10、Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.1) 2x2 3x 20xyo0.52Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.2) 3x2 6
11、x 2xyoxyo3x2 6x 20Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.3) 4x2 4x 10xo0.5yEvaluation only.Evaluation only. C
12、reated with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.4) x2 2x 3xoyx2 2x 30xoyEvaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client
13、Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.1) 2x2 3x 200.52+- 解二： 2x2 3x 20 (2x+1)(x2)0(2x+1) 0(2x+1) 0 (x2) 0(x2) 0或Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Cli
14、ent Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.解：由数轴标根法（如图），得01-13+-+-1x0 或 1x3Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspos
15、e.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.x2或x-1-1x2xyo-12y分析1：用数形结合的方法解之（如图）分析2：函数与轴的两个交点横坐标-1和2，实为 方程 y=x2 +bx +c 的两根，由此可求 b,c 即 -12 =-b， (-1)2 =c 由此，问题转化成解不等式：Evaluation only.Evaluation only. Created with Aspose.Slides for
16、.NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.Copyright 2004-2011 Aspose Pty Ltd.Rxoy分析：用数形结合的方法解之（如图）Evaluation only.Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0.Created with Aspose.Slides for .
