2022年因式分解法解一元二次方程 .pdf

学习好资料欢迎下载教案：因式分解法解一元二次方程开福区植基中学陈雪晴 2016 09 目标：熟练地运用因式分解法解一元二次方程重点：用因式分解法解一元二次方程难点：学生通过比较解一元二次方程的多解方法，感悟用因式分解法解题的简便。一、复习引入（学生活动）（1）2X2+X=0(用配方法)（2）5X2+6X=0（用公式法）二、探索新课（1）学生活动分解因式1、2X2+X=X（2X+1）2、5X2+6X=（5X+6）（2）学生思考若 X（2X+1）=0（1）X（5X+6）=0（2）师问：在（1）中，X与（2X+1）有怎么样关系？在（2）中 X 与（5X+6）有怎样的关系？3、引入新课（板书课题）因式分解法解一元二次方程（1）X2+2X=0（2）4X2-144=0 学习好资料欢迎下载（3）3X2-12X=-12(4)X2-3X+2=0(4)例题 1：教师板书解题过程解：X2+2X=0 X（X+2）=0 （提公因式）X=0或 X+2=0（化为两个一元一次方程）X1=0或 X2=-2（5）师讲授等式左边因为两个因乘积要等于0，所以至少要其中一个因式为0 也就是（1）X=0或 2X+1=0，所以 X1=0 X2=-2 问题：以上方是如何实现降次的？（师生共同总结）把一元二次方程变成两个一元一次方程来降次解（2）X2-144=0 问题：方程左边可以变成哪两个因式的积的形式？（学生演示）1、（2X-12）（2X+12）=0 2、4（X2-36）=0（提公因式）（X-6）（X+6）=0（平方差公式）（板书解题过程）4X2-144=0 4（X2-36）=0（X-6）（X+6）=0 文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5文档编码:CI6D6M3Q2O1 HU6N7U10T6W7 ZN5T1R4V10B5学习好资料欢迎下载X-6=0 或 X+6=0 X1=6 或 X2=-6 师生共同总结，这种题型利用平方差公式分解成（a+b）(a-b)=0的形式来解方程达到降次的目的。例题 3 解方程3X2-12X=12（学生板书）3 X-12X+12=0 3（X2-4X+4）=0（提公因式）X2-4X+4=0（X+2）2=0（完全平方公式）X-2=0 即 X1=X2=2 总结：此类题型利用完全平方公式变成（ab）2=0 形式降次来解一元二次解例题 4 解方程X2-2X+2=0 1、引导学生认真观察，能用以上3 种方法来解决此题吗？2、怎样把等式左边变成积的形式呢？1-2 1-1 1X1=1 文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 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HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8学习好资料欢迎下载-2X(-1)=2 1X(-1)+1(-2)=-3 像这种题型，不能提公因式或公因数也不能直接用公式的，试着用这种“+”字交叉法。注：分解二次项系数，分解常数交叉相乘的积的和等于一次项系数。所以 X2+(a+b)x+ab=0 的形式都可以用“十”字交叉法所以 X2-3X+1=0（X-2）（X-1）=0 X-2=0 或 X-1=0 X1=2 X2=1 三、交流探讨回顾小结本节课的内容是用因式分解的方法来解一元二次方程，步凑是：1、提公式或公因数2、用公式3、十字交叉方法的变在ab=0的形式。达到降次变成两个一元一次方程的形式，从降次而使问题得以解决。四、学生质疑，你对这节课有哪些疑点问点？五、应用扩展解方程（2X-1）2=（3-X）23（X-1）-2（X-1）=0 六、作业书本知识P17、6 文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 HT2W10S5T3Z9 ZU8V4A8R4N8文档编码:CK10W3A1K7I4 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