20.1认识二次根式 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20.1《认识二次根式》专项练习##一、知识点梳理一般地(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)形如$\boldsymbol{\sqrt{a}(a\ge0)}$的式子叫做二次根式。其中$a$叫做被开方数(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)“$\sqrt{\quad}$”称为二次根号。二次根式有两个必备条件：一是含有二次根号；二是被开方数必须是非负数。也就是说(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)只有当被开方数大于或等于0时(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)二次根式才有意义。由此衍生出核心考点：根据二次根式有意义(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)求字母的取值范围。二次根式具备双重非负性：①被开方数$a\ge0$；②二次根式本身$\sqrt{a}\ge0$。双重非负性是本章最常考的性质(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)常结合绝对值、平方数的非负性综合出题。因为一个数的绝对值、实数的平方、二次根式的结果都大于等于0(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)如果几个非负数相加等于0(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)则每一项都必须等于0。同时要区分$\sqrt{a}$、$-\sqrt{a}$、$\pm\sqrt{a}$的含义(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)$\sqrt{a}$表示$a$的算术平方根(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)结果一定是非负数；$-\sqrt{a}$表示算术平方根的相反数；$\pm\sqrt{a}$表示$a$的平方根。注意二次根式$\sqrt{a}$（$a\ge0$）是一个整体(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)它代表一个非负实数。##二、基础选择题1.下列式子中(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)属于二次根式的是（）A.$\sqrt{-3}$B.$\sqrt{5}$C.$\sqrt[3]{4}$D.$x$2.若二次根式$\sqrt{x-2}$在实数范围内有意义(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)则$x$的取值范围是（）A.$(x>2)$B.$x\ge2$C.$(x<2)$D.$x\le2$3.若$\sqrt{a+1}+|b-2|=0$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)则$(a+b)$的值为（）A.$-1$B.$0$C.$1$D.$2$4.下列说法正确的是（）A.$\sqrt{-2}$是二次根式B.当$(a<0)$时(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)$\sqrt{a}$也有意义C.$\sqrt{4}$是二次根式D.二次根式的值一定是正数5.要使$\sqrt{3-x}+\dfrac{1}{\sqrt{x-1}}$同时有意义(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)则$x$的取值范围是（）A.$1<x\le3$B.$x\ge3$C.$(x>1)$D.$1\lex\le3$##三、填空题1.当$(x=)$******时(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)二次根式$\sqrt{x+5}$取到最小值(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)最小值为******。2.若式子$\dfrac{\sqrt{2x-1}}{x-3}$有意义(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)则$x$的取值范围是________。3.已知$\sqrt{m-4}+(n+2)^2=0$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)则$m-n=$________。4.判断：$\sqrt{0}$______二次根式（填“是”或“不是”）。5.若$\sqrt{(x-1)^2}=1-x$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)则$x$的取值范围是________。##四、解答题1.求下列各式中字母的取值范围。（1）$\sqrt{2-3x}$；（2）$\dfrac{\sqrt{x+2}}{x}$；（3）$\sqrt{x^2+1}$2.已知实数$(x,y)$满足$\sqrt{x-3}+|y+1|=0$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)求代数式$x-2y$的值。3.已知$a$为实数(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)求代数式$\sqrt{a+2}+\sqrt{-a-2}$的值。4.拓展探究：若$\sqrt{(2a-1)^2}=1-2a$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)求$a$的取值范围。##五、参考答案与解析###选择题1.B。解析：A被开方数负数(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)无意义；C是立方根；D不含根号。2.B。解析：$x-2\ge0$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)解得$x\ge2$。3.C。解析：两个非负数相加等于0(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)则$\sqrt{a+1}=0(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)|b-2|=0$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)$a=-1(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)b=2$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)$(a+b)=1$。4.C。解析：负数不能开平方；$\sqrt{0}=0$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)二次根式的值可以是0。5.A。解析：$\begin{cases}3-x\ge0\x-1>0\end{cases}$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)解得$1<x\le3$。###填空题1.$-5$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)$0$；解析：$\sqrt{x+5}\ge0$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)最小值为0(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)此时$(x+5=0)$。2.$x\ge\dfrac12$且$x\neq3$；解析：被开方数≥0(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)分母≠0。3.$6$；解析：$m=4(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)n=-2$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)$m-n=4-(-2)=6$。4.是；解析：$\sqrt{0}$满足定义(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)$(a=0≥0)$。5.$x\le1$。###解答题1.（1）$2-3x\ge0$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)$x\le\dfrac23$（2）$\begin{cases}x+2\ge0\x\neq0\end{cases}$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)$x\ge-2$且$x\neq0$（3）$x^2+1$恒大于0(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)$x$取全体实数。2.解：$\because\sqrt{x-3}\ge0(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)|y+1|\ge0$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)且$\sqrt{x-3}+|y+1|=0$$\thereforex-3=0(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)y+1=0$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)得$(x=)3(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)y=-1$代入$x-2y=3-2\times(-1)=5$。3.解：$\begin{cases}a+2\ge0\-a-2\ge0\end{cases}$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)$\begin{cases}a\ge-2\a\le-2\end{cases}$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)$\thereforea=-2$。原式$=\sqrt0+\sqrt0=0$。4.解：$\sqrt{(2a-1)^2}=|2a-1|=1-2a$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)说明$2a-1\le0$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)解得$a\le\dfrac12$。##六、易错总结1.求取值范围时分母不能为0(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)很多同学容易忽略分母限制条件。2.二次根式结果≥0(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)不是一定大于0(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)$\sqrt{0}=0$也是二次根式。3.利用非负性解题(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)多个非负数相加为0(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)每一项单独等于0(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)这是高频考点。4.$\sqrt{a^2}=|a|$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)不是直接等于$a$(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)需要分类讨论符号(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)是本节最容易出错的点。（全文：1442字）要不要我再单独出一份不带答案的空白试卷版(\sqrt{a+1}=0(a=-1(m=4(\because\sqrt{x-3}\ge0(\thereforex-3=0(x=3，y=-1)y+1=0)|y+1|\ge0)n=-2)b=2)|b-2|=0)方便你直接打印做题？华东师大版（新教材）数学九年级上册20.1认识二次根式授课教师：.

班

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9年级（

）班.

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2026/10/10

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。第二十章二次根式123理解二次根式的概念，会判断一个式子是否为二次根式。掌握二次根式有无意义的条件，领会数学分类讨论思想.理解二次根式的性质，能运用其进行化简计算.在八年级上册第10章，我们学习了平方根与算术平方根。

回顾训练计算下列各数的平方根和算术平方根

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