21.2.2配方法 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21.2.2《配方法》专项练习（全文1444字）##一、知识点梳理配方法\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)本质是把一元二次方程通过变形\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)转化成$\boldsymbol{(x+m)^2=n(n\ge0)}$的形式\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)再利用直接开平方法求解。核心思想是**转化**\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)将二次方程转化为已经学过的直接开平模型。配方法解一元二次方程$ax^2+bx+c=0(a\neq0)$的标准步骤：1.**化1**：二次项系数化为1\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)方程两边同除以$a$；2.**移项**：常数项移到方程右边\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)只留含未知数的项在左边；3.**配方**：方程左右两边同时加上**一次项系数一半的平方**\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)使左边成为完全平方式；4.**写成平方形式**：左边写成$(x+m)^2$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)合并右边常数；5.**开方求解**：若右边$n\ge0$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)直接开平方解方程；若$\(n<0\)$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)方程无实数根。配方的关键公式：$x^2+px+\boldsymbol{\left(\dfrac{p}{2}\right)^2}=\left(x+\dfrac{p}{2}\right)^2$。注意：只有二次项系数为1时\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)才能直接加上一次项系数一半的平方；二次项系数不为1\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)必须先除以系数化为1再配方。配方法还可以用来求二次代数式的最值、证明代数式恒大于或小于某个数。本节高频易错点：①二次项系数不为1时\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)忘记先化1\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)直接加一次项系数一半的平方；②配方时\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)只在左边加常数\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)右边不加\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)破坏等式；③计算“一次项系数一半的平方”时符号出错；④移项时不变号；⑤开平方漏掉负号；⑥配方完成后\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)判断右边常数小于0\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)仍然强行求根。##二、基础选择题1.将$x^2-6x$配方\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)需要加上（）A.$3$B.$-3$C.$9$D.$-9$2.用配方法解方程$x^2-4x-5=0$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)配方后得到（）A.$(x-2)^2=9$B.$(x-2)^2=1$C.$(x+2)^2=9$D.$(x-4)^2=21$3.把方程$2x^2-4x-1=0$化为$(x+m)^2=n$形式\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)正确是（）A.$(x-1)^2=\dfrac32$B.$(x-1)^2=\dfrac12$C.$(x-2)^2=5$D.$(x-2)^2=\dfrac52$4.代数式$x^2+2x+3$的最小值是（）A.$2$B.$3$C.$4$D.$5$5.方程$x^2-2x+3=0$配方后\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)下列说法正确的是（）A.有两个不相等实数根B.有两个相等实数根C.无实数根D.根为$x_1=3,x_2=-1$##三、填空题1.$x^2+8x+$____$\(=(x+\)$____$)^2$。2.方程$x^2+4x=1$配方后为________________。3.用配方法解方程$x^2-2x-1=0$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)配方后：$(x-\underline{\quad})^2=\underline{\quad}$。4.将$3x^2-6x+1=0$二次项系数化为1得：________________。5.若$x^2-10x+m$是完全平方式\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)则$\(m=\)$________。##四、解答题1.用配方法解下列方程：（1）$x^2-8x+7=0$；（2）$x^2+6x-16=0$2.用配方法解方程：（1）$2x^2-4x-6=0$；（2）$3x^2+6x-1=0$3.利用配方法求代数式$x^2-4x+5$的最小值\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)并写出此时$x$的值。4.拓展探究：求证：无论$x$取任何实数\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)代数式$2x^2-8x+11$的值恒大于0。##五、参考答案与解析###选择题1.C。解析：一次项系数$-6$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)一半是$-3$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)平方是$9$。2.A。解析：移项$x^2-4x=5$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)两边加4\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)$x^2-4x+4=9$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)即$(x-2)^2=9$。3.A。解析：两边除以2：$x^2-2x=\dfrac12$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)配方两边加1\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)$(x-1)^2=\dfrac32$。4.A。解析：$x^2+2x+3\(=(x+\)1)^2+2$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)$(x+1)^2\ge0$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)最小值2。5.C。解析：配方$(x-1)^2=-2$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)右边负数\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)无实数根。###填空题1.$16\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)4$2.$(x+2)^2=5$；解析：两边加4\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)$x^2+4x+4=1+4$。3.$1\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)2$；解析：$x^2-2x+1=1+1$。4.$x^2-2x+\dfrac13=0$5.$25$；解析：一次项系数一半$-5$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)平方25。###解答题1.（1）移项：$x^2-8x=-7$配方：$x^2-8x+16=-7+16$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)$(x-4)^2=9$开方：$x-4=\pm3$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)$x_1=7\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)x_2=1$（2）移项：$x^2+6x=16$配方：$x^2+6x+9=16+9$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)$(x+3)^2=25$开方：$x+3=\pm5$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)$x_1=2\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)x_2=-8$2.（1）方程两边除以2：$x^2-2x-3=0$移项：$x^2-2x=3$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)配方$x^2-2x+1=4$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)$(x-1)^2=4$开方：$x-1=\pm2$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)$x_1=3\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)x_2=-1$（2）两边除以3：$x^2+2x-\dfrac13=0$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)移项$x^2+2x=\dfrac13$配方两边加1：$x^2+2x+1=\dfrac13+1$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)$(x+1)^2=\dfrac43$开方：$x+1=\pm\dfrac{2\sqrt3}{3}$$x_1=-1+\dfrac{2\sqrt3}{3}\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)x_2=-1-\dfrac{2\sqrt3}{3}$3.解：$x^2-4x+5=(x^2-4x+4)-4+5=(x-2)^2+1$$\because(x-2)^2\ge0$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)$\therefore(x-2)^2+1\ge1$当$\(x=2\)$时\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)代数式取得最小值$1$。4.证明：$2x^2-8x+11=2(x^2-4x)+11=2(x^2-4x+4-4)+11$$=2[(x-2)^2-4]+11=2(x-2)^2-8+11=2(x-2)^2+3$$\because(x-2)^2\ge0$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)$\therefore2(x-2)^2\ge0$\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)$2(x-2)^2+3\ge3>0$故无论$x$取任何实数\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)代数式的值恒大于0。##六、易错总结配方法最核心的坑：**二次项系数不为1\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)必须先同除以二次项系数\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)再配方**。很多同学不化1直接加一次项系数一半的平方\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)整道题全部错误。配方是等式变形\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)左边加多少\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)右边必须同步加上相同的数\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)等式才能保持成立\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)只加左边不加右边是高频错误。配方时\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)一次项带负号\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)不要漏掉符号\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)计算一半的平方永远是正数。配方完成后\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)右边常数小于0\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)直接判定方程没有实数根\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)不能继续开平方。配方法不仅用来解方程\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)还常用来求二次多项式最值、证明代数式恒正或恒负\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)这类题型在考试里很常见。求最值的思路：把代数式配成完全平方加常数\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)利用平方的非负性判断。解方程做完可以代回检验\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)验证左右两边相等。配方法是推导一元二次方程求根公式的基础\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)学好配方法\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)才能理解后面的公式法。做题时注意区分：解方程的配方\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)和整式中对二次三项式配方\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)二者的异同点\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)整式配方不需要等式\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)只变形代数式；方程配方是等式\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)两边同时变形。（全文1444字）下一节：**21.2.3公式法**\(16\(1\(x_1=7\(x_1=2\(x_1=3\(x_1=-1+\dfrac{2\sqrt3}{3}，x_2=-1-\dfrac{2\sqrt3}{3}\)x_2=-1\)x_2=-8\)x_2=1\)2\)4\)继续同格式1444字专项练习吗？华东师大版（新教材）数学九年级上册21.2.2配方法授课教师：.

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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123掌握配方法的推导过程，掌握配方法解一元二次方程的步骤.熟练应用配方法解一元二次方程.能根据一元二次方程的特点，灵活运用配方法.在配方法的应用过程中体会“转化”的思想，掌握一些转化的技能.直接开平方法和因式分解法解一元二次方程的应用场景？

03新知探究探究配方法解一元二次方程为此，通常设法在方程两边都加上一个适当的数，使左边变成一个含有未知数的完全平方式(右边是一个常数).那么，本题中，要把x2+2x=5的左边配成完全平方式，这个“适当的数”是什么呢?【例】解方程：x2+2x=5.【思考】要用直接开平方法求解，首先希望能将方程化为（)2=a的形式，那么，怎么实现呢?03新知探究探究配方法解一元二次方程【例】解方程：x2+2x=5.将一次项2x改写成2·x·1，得x2+2·x·1=5由此（a+b）2=a2+2ab+b2可以看出，为使左边成为完全平方式，只需在方程两边都加上12即：x2+2·x·1+12=5+12，

（x+1）2=6现在你会解这个方程了吗？03新知探究探究配方法解一元二次方程【例】解方程：x2+2x=5.解：原方程两边都加上1，得x2+2x+1=6，即(x+1)2=6.直接开平方，得x+1=所以x=-1即x1=，x2=.总结归纳上面的解方程，是通过将方程简单变形，将左边配成一个含有未知数的完全平方式，右边是一个非负常数，从而可以直接开平方求解.这种解一元二次方程的方法叫做配方法.x2+2x=5(x+1)2=6.一元二次方程完全平方式非负常数03新知探究x2+2x+___=(________)2x2-2x+___=(________)2x2+4x+___=(________)2x2-4x+___=(________)2x2+6x+___=(________)2x2-6x+___=(________)2x2+10x+___=(________)2x2-10x+___=(________)2

1x+11x-14x+24x-29x+39x-325x+525x-5【做一做】添上一个适当的数，使下列的多项式成为一个完全平方式03新知探究配方的关键：对二次项系数为1的一元二次方程x2+bx=c配方，只需在方程两边都加上一次项系数一半的平方。常数项是一次项系数一半的平方【思考】在用配方法解二次项系数是1的一元二次方程时，添上的常数项与一次项系数之间存在着什么样的关系？03新知探究【例5】用配方法解方程：（1）x2-4x+1=0.（2）4x2-12x-1=0解：（1）原方程可化为x2-4x=-1，配方（两边都加上4），得x2-2·x·2+22=-1+22，即(x-2)2=3.直接开平方，得x-2=所以x1=，x2=.总结归纳配方法解二次项系数为1的一元二次方程的基本步骤：(1)移项：把常数项移到方程的右边；(2)配方：方程两边同时加上一次项系数一半的平方；(3)开方：根据平方根的意义,方程两边开平方;(4)求解：解一元一次方程;(5)定解：写出原方程的解.知识点1

用配方法解二次项系数为1的一元二次方程

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