高二數(shù)學(xué)一元二不次等式及其解法

一元二次不等式的解法ax2+bx+c＞0ax2+bx+c＜0(a＞0)xyox1x2xyox1x2xyox1x2xyox1x2xyox1x2xyox1x2Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.解一元一次不等式時(shí)，你們知道應(yīng)具備什么知識嗎？1)若a＞d，則a＋c＞d＋c2)若a＞d且c＞0，則ac＞dc3)若a＞d且c＜0，則ac＜dc不等式的基本性質(zhì)：Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.還有一種數(shù)學(xué)方法可以解不等式，你們知道嗎？數(shù)形結(jié)合在解不等式中起著非常優(yōu)越的作用！先看它在解一元一次不等式的應(yīng)用吧！xyo3.5和①解方程2x-7=0②作函數(shù)y=2x-7的圖象③解不等式:2x-7＞02x-7＜0數(shù)形結(jié)合(y＞0)(y＜0)Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.你們現(xiàn)在可以用數(shù)形結(jié)合解一元二次不等式了嗎？①解方程x2

–x–6=0②作函數(shù)y=x2

–x–6的圖象③解不等式:-23-6xyox2

–x–6＞0x2

–x–6＜0Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.由特殊到一般一元二次不等式解法方程：ax2+bx+c=0的解情況函數(shù)：y=ax2+bx+c的圖象不等式的解集ax2+bx+c＞0ax2+bx+c＜0a＞0xyox1x2xox0yxoy當(dāng)⊿＞0時(shí)，方程有兩不等的根：x1，x2當(dāng)⊿＝0時(shí)，方程有一根：

x0當(dāng)⊿＜0時(shí)，方程無解{x∣x＜x1

或x＞x2}{x∣x≠x0}R{x∣x1＜x＜x2}??Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.解下列的不等式:1)2x2－3x－2＞02)－3x2＋6x＞23)4x2－4x＋1＞04)－x2＋2x－3＞0Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.解不等式中的數(shù)形結(jié)合注意：由右到左，先正后負(fù)2)數(shù)軸上標(biāo)根后，必須遵循的順序填入+、-號1)每個(gè)因式中的x的系數(shù)必須是正數(shù)－0.52++-(2x+1)(x－2)＞0——可解高次不等式1、函數(shù)圖象法2、數(shù)軸標(biāo)根法Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.練習(xí)，練習(xí)，再練習(xí)！Rx＞2或x＜-1-1＜x＜2-12-2-2≤a≤61Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.一元二次不等式的解法學(xué)會了嗎？回去好好總結(jié)消化哦！還有不明白的地方一定要找老師弄個(gè)明白！多多訓(xùn)練不可少！88:)88:)88:)88:)88:)88:)Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.解下列的不等式:1)2x2－3x－2＞0xyo－0.52Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.解下列的不等式:2)－3x2＋6x＞2xyoxyo－3x2＋6x－2＞0Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.解下列的不等式:3)4x2－4x＋1＞0xo0.5yEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.解下列的不等式:4)－x2＋2x＞3xoy－x2＋2x－3＞0xoyEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.解下列的不等式:1)2x2－3x－2＞0－0.52++-解二：2x2－3x－2＞0(2x+1)(x－2)＞0(2x+1)＞0(2x+1)＜0(x－2)＞0(x－2)＜0或Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.解：由數(shù)軸標(biāo)根法（如圖），得01-13+-++--1＜x＜0或1＜x＜3Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.練習(xí)，練習(xí)，再練習(xí)！x＞2或x＜-1-1＜x＜2xyo-12y分析1：用數(shù)形結(jié)合的方法解之（如圖）分析2：函數(shù)與軸的兩個(gè)交點(diǎn)橫坐標(biāo)-1和2，實(shí)為方程y=x2+bx+c

的兩根，由此可求b,c即-1＋2=-b，(-1)×2=c由此，問題轉(zhuǎn)化成解不等式：Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.練習(xí)，練習(xí)，再練習(xí)！Rxoy分析：用數(shù)形結(jié)合的方法解之（如圖）Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.練習(xí)，練習(xí)，再練習(xí)！-12-2分析：不等式的解中的和其實(shí)就是方程的兩個(gè)根Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile.Copyright2004-2011AsposePtyLtd.練習(xí)，練習(xí)，再練習(xí)！-2≤a≤6分析：對于函數(shù)的圖象依題意，應(yīng)可作圖如下xoy要使圖象處于此位置，則即Evaluationonly.Createdwith