21.2.1.2 直接开平方法和因式分解法 课件 -2026-2027学年华东师大版数学九年级上册

#华东师大版九年级上册数学21.2.1.2《直接开平方法和因式分解法（2）》专项练习（全文1444字）##一、知识点梳理本节在直接开平方法基础上\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)学习**因式分解法**解一元二次方程。因式分解法核心原理：如果两个因式的积等于0\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)那么至少其中一个因式等于0\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)即若$\(a\(b=0\)\)$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)则$\(a=0\)$或$\(b=0\)$。这种方法适合方程右边为0\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)左边可以分解成两个一次因式乘积的一元二次方程。因式分解法解题步骤：第一步移项\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)把方程所有项移到左边\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)使方程右边等于0；第二步\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)对左边多项式因式分解\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)常用方法有提公因式法、平方差公式、完全平方公式；第三步\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)令每一个因式分别等于0\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)得到两个一元一次方程；第四步\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)解一元一次方程\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)得到一元二次方程的两个根。对比两种解法：直接开平方法适合形如$x^2=p$或$(mx+n)^2=p$的方程；因式分解法适合容易分解因式的方程\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)如$x^2+b\(x=0\)$（提公因式）、$a^2x^2-b^2=0$（平方差）。部分方程两种方法都可以解\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)可灵活选择。重要注意点：解方程时\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)**不能随意在方程两边同时除以含有未知数的代数式**\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)否则会丢失一个根。例如$x(x-2)=3x$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)不能直接两边同除以$x$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)否则漏掉$\(x=0\)$这个根。移项是前提\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)必须化为“左边因式\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)右边等于0”的形式。本节高频易错点：①没有移项\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)右边不为0就分解因式；②方程两边直接约去含未知数因式\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)造成丢根；③因式分解不彻底；④使用平方差、完全平方公式时符号出错；⑤混淆“或”与“且”\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)两个因式是只要一个为0即可；⑥书写根的时候\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)相等的根写成$x_1=x_2$；⑦分不清什么时候用直接开平方法\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)什么时候用因式分解法。##二、基础选择题1.解方程$x(x-3)=0$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)正确结果是（）A.$\(x=0\)$B.$\(x=3\)$C.$x_1=0,x_2=3$D.$x_1=0,x_2=-3$2.方程$x^2-16=0$用因式分解法\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)正确分解是（）A.$(x-4)^2=0$B.$(x+4)(x-4)=0$C.$x(x-16)=0$D.$(x-8)(x+2)=0$3.解方程$\((\(x+1\))\)^2=2\((\(x+1\))\)$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)两边不能直接除以$\((\(x+1\))\)$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)原因是（）A.$\(x+1\)$可能等于0\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)会丢根B.$\(x+1\)$是正数C.$\(x+1\)$是负数D.等式不能做除法4.方程$(2x-1)^2=9$的根是（）A.$\(x=2\)$B.$x=-1$C.$x_1=2,x_2=-1$D.$x_1=4,x_2=-2$5.下列方程适合用因式分解法求解的是（）A.$x^2-4x-1=0$B.$(x-2)^2=12$C.$x^2-6\(x=0\)$D.$2x^2+3x=7$##三、填空题1.方程$x^2-5\(x=0\)$因式分解得$x(x-5)=0$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)根为____________。2.方程$(x-2)^2=3(x-2)$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)移项后因式分解为____________。3.方程$4x^2-9=0$的根是________________。4.若$(x-3)(x+5)=0$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)则$x_1=$____\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)$x_2=$____。5.方程$x^2-4x+4=0$的根是____________。##四、解答题1.选用合适方法解方程（直接开平方法或因式分解法）：（1）$x^2-7\(x=0\)$；（2）$x^2-25=0$；（3）$(x+3)^2=16$；（4）$(x-2)^2=3(x-2)$2.解方程：（1）$4(x-1)^2-9=0$；（2）$x^2-6x+9=0$3.已知一元二次方程$x^2+kx-12=0$有一个根是$\(x=3\)$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)求$k$的值与方程另一个根。4.拓展探究：解方程$(2x-1)^2=(x+2)^2$。##五、参考答案与解析###选择题1.C。解析：两个因式乘积为0\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)$\(x=0\)$或$x-3=0$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)两根$\(0,3\)$。2.B。解析：平方差公式$x^2-4^2=(x+4)(x-4)$。3.A。解析：$\(x+1\)$有可能等于0\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)等式两边不能除以0\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)直接除会丢失$x=-1$这个根。4.C。解析：直接开平方\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)$2x-1=\pm3$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)解得$x_1=2,x_2=-1$。5.C。解析：提取公因式$x$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)$x(x-6)=0$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)适合因式分解。###填空题1.$x_1=0,x_2=5$2.$(x-2)(x-5)=0$；解析：移项$(x-2)^2-3(x-2)=0$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)提公因式$(x-2)(x-2-3)=0$。3.$x_1=\dfrac32,x_2=-\dfrac32$；解析：$(2x+3)(2x-3)=0$。4.$3\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)-5$5.$x_1=x_2=2$；解析：完全平方$(x-2)^2=0$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)两个相等实数根。###解答题1.（1）$x(x-7)=0$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)$x_1=0,x_2=7$（2）$(x+5)(x-5)=0$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)$x_1=-5,x_2=5$（3）直接开方：$x+3=\pm4$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)$x_1=1\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)x_2=-7$（4）移项：$(x-2)^2-3(x-2)=0$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)$(x-2)(x-5)=0$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)$x_1=2,x_2=5$2.（1）$4(x-1)^2=9$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)$(x-1)^2=\dfrac94$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)$x-1=\pm\dfrac32$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)$x_1=\dfrac52\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)x_2=-\dfrac12$（2）$(x-3)^2=0$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)$x_1=x_2=3$3.解：把$\(x=3\)$代入方程\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)$9+3k-12=0$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)解得$\(k=1\)$。方程为$x^2+x-12=0$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)因式分解$(x+4)(x-3)=0$\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)另一根$x=-4$。4.解：移项：$(2x-1)^2-(x+2)^2=0$平方差分解：$[(2x-1)+(x+2)][(2x-1)-(x+2)]=0$$(3\(x+1\))(x-3)=0$$3\(x+1\)=0$或$x-3=0$$x_1=-\dfrac13\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)x_2=3$##六、易错总结因式分解法第一步必须移项\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)把方程右边化为0\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)很多同学不等右边为0就分解\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)直接出错。最容易丢分的陷阱：方程两边不能同时除以含有未知数的式子\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)因为这个式子有可能等于0\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)会丢掉一个根\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)必须移项提公因式。因式分解要熟练掌握提公因式、平方差、完全平方公式\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)分解一定要分解到不能再分解为止。区分两种解法：有完全平方等于常数\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)优先直接开平方法；多项式容易分解成两个一次因式\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)优先因式分解法。做题时注意“或”的含义\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)两个因式只需要一个等于0即可\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)不是同时等于0。遇到完全平方式等于0\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)结果要写两个相等实数根\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)不能只写一个解。求出根之后可以代入原方程检验\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)验证左右两边相等。当方程两边都是平方形式\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)既可以直接开平方\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)也可以移项后用平方差因式分解\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)两种方法都可以\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)可选择计算简单的一种。计算时注意符号\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)去括号、移项时符号极易出错。本节是一元二次方程基础解法\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)熟练选择合适方法\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)能大幅提升解方程速度\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)为后面配方法、公式法打好基础。（全文1444字）下一节：**21.2.2配方法**\(3\(x_1=1\(x_1=\dfrac52\(x_1=-\dfrac13，x_2=3\)x_2=-\dfrac12\)x_2=-7\)-5\)继续同规格1444字专项练习吗？华东师大版（新教材）数学九年级上册21.2.1.2直接开平方法和因式分解法（2）授课教师：.

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2026/10/11

学子如同向日葵向往暖阳，满心热爱渴求知识。在书页间汲取力量，于思索中追逐光亮，不惧学习路上的磨砺，心怀憧憬奋力向上，向着理想不断成长。第21章一元二次方程02新知导入【思考】直接开平方法、因式分解法的适用标准方程形式是什么？直接开平方法基本形式：x2=p(p≥0)或(ax+b)2=p(p≥0)。适用条件：方程的一边是含有未知数的代数式的平方，另一边是非负数。若p<0，则方程无实数根。02新知导入【思考】直接开平方法、因式分解法的适用标准方程形式是什么？因式分解法基本形式：方程右边为0，左边为两个一次因式的乘积，即ax2+bx+c=0且左边能成功因式分解(如提公因式、平方差、完全平方、十字相乘等)为(px+q)(mx+n)=0的形式。适用条件：方程必须整理为一般式，且右边必须化为0，左边易于分解。03新知探究【例3】解下列方程:(1)(x+1)2-4=0;(2)12(2-x)2-9=0.【分析】两个方程都可以通过简单的变形，化为（)2=a(a≥0)的形式，用直接开平方法求解.03新知探究【例3】解下列方程:(1)(x+1)2-4=0;(2)12(2-x)2-9=0.解：(1)原方程可以变形为(x+1)2=4.直接开平方，得x+1=±2.所以x1=1，x2=-3.还有其他解法吗？03新知探究【例3】解下列方程:(1)(x+1)2-4=0;(2)12(2-x)2-9=0.解：(1)原方程可以变形为(x+1)2-22=0.方程左边分解因式，得（x+1+2）（x+1-2）=0.（x+3）（x-1）=0所以x1=1，x2=-3.03新知探究【例3】解下列方程:(1)(x+1)2-4=0;(2)12(2-x)2-9=0.解：(2)原方程可以变形为(2-x)2=

.直接开平方，得2-x=±.所以x1=，x2=.这个题目能用因式分解法吗？03新知探究【例3】解下列方程:(1)(x+1)2-4=0;(2)12(2-x)2-9=0.解：(2)原方程可以变形为(2-x)2-

=0.方程左边分解因式，得（2-x+）（2-x-）=0.所以x1=1，x2=-3.