《因式分解公式法》PPT课件

八年级数学（上册）Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.课前小测：1.选择题：1)下列各式能用平方差公式分解因式的是（）A.4X²+y²B.4x-(-y)²C.-4X²-y³D.-X²+y²2)-4a²+1分解因式的结果应是（）A.-(4a+1)(4a-1)B.-(2a–1)(2a–1)C.-(2a+1)(2a+1)D.-(2a+1)(2a-1)2.把下列各式分解因式：1）18-2b²2)x4–1DD1)原式=2（3+b)(3-b)2)原式=(x²+1)(x+1)(x-1)Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.因式分解的基本方法2运用公式法把乘法公式反过来用,可以把符合公式特点的多项式因式分解,这种方法叫公式法.(1)平方差公式：a2-b2=(a+b)(a-b)(2)完全平方公式：a2+2ab+b2=(a+b)2a2-2ab+b2=(a-b)2Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.平方差公式反过来就是说：两个数的平方差，等于这两个数的和与这两个数的差的积a²-b²=(a+b)(a-b)因式分解平方差公式：(a+b)(a-b)=a²-b²整式乘法Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.将下面的多项式分解因式1)m²-162)4x²-9y²m²-16=m²-4²=(m+4)(m-4)a²-b²=(a+b)(a-b)4x²-9y²=(2x)²-(3y)²=(2x+3y)(2x-3y)Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.例1.把下列各式分解因式（1）16a²-1(2)4x²-m²n²(3)—x²-—y²925116(4)–9x²+4解：1）16a²-1=(4a)²-1=(4a+1)(4a-1)解：2）4x²-m²n²=(2x)²-(mn)²=（2x+mn)(2x-mn)Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.例2.把下列各式因式分解1)(x+z)²-(y+z)²2)4(a+b)²-25(a-c)²3)4a³-4a4)(x+y+z)²-(x–y–z)²5)—a²-212解：1.原式=[(x+z)+(y+z)][(x+z)-(y+z)]=(x+y+2z)(x-y)解：2.原式=[2(a+b)]²-[5(a-c)]²=[2(a+b)+5(a-c)][2(a+b)-5(a-c)]=(7a+2b-5c)(-3a+2b+5c)解：3.原式=4a(a²-1)=4a(a+1)(a-1)解：4.原式=[(x+y+z)+(x-y-z)]×[(x+y+z)-(x-y-z)]=2x(2y+2z)=4x(y+z)Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.巩固练习：1.选择题：1)下列各式能用平方差公式分解因式的是（）A.4X²+y²B.4x-(-y)²C.-4X²-y³D.-X²+y²2)-4a²+1分解因式的结果应是（）A.-(4a+1)(4a-1)B.-(2a–1)(2a–1)C.-(2a+1)(2a+1)D.-(2a+1)(2a-1)2.把下列各式分解因式：1）18-2b²2)x4–1DD1)原式=2（3+b)(3-b)2)原式=(x²+1)(x+1)(x-1)Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.2ab2ab222aabb222aabb完全平方公式Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.2ab2ab222aabb222aabb现在我们把这个公式反过来很显然，我们可以运用以上这个公式来分解因式了，我们把它称为“完全平方公式”Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.我们把以上两个式子叫做完全平方式222aabb222aabb“头”平方,“尾”平方,“头”“尾”两倍中间放.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.判别下列各式是不是完全平方式2222222224232221乙乙甲甲BABAyxyx是是是是Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.完全平方式的特点：1、必须是三项式222首首尾尾2、有两个平方的“项”3、有这两平方“项”底数的2倍或-2倍222aabb222aabbEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.下列各式是不是完全平方式22222222222122234446154624ababxyxyxxyyaabbxxaabb是是是否是否Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.请补上一项，使下列多项式成为完全平方式222222224221_______249_______3______414_______452______xyabxyabxxy2xy12ab4xyab4yEvaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.2ab2ab222aabb222aabb我们可以通过以上公式把“完全平方式”分解因式我们称之为：运用完全平方公式分解因式Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.例题：把下列式子分