3.2.2二次根式的除法 课件 -2026-2027学年湘教版数学八年级上册

#湘教版数学八年级上册\(\boldsymbol{3}\).2.2二次根式的除法专项练习##知识点回顾二次根式除法法则：$$\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)b>0)}$$文字表述：两个二次根式相除\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)把被开方数相除\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)根指数不变。逆用（商的算术平方根）：$$\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)b>0)}$$解题要点：1.公式条件：$a\ge0\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)\boldsymbol{b>0}$\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)分母不能为0；2.系数相除、根式分开运算：$\dfrac{m\sqrt{a}}{n\sqrt{b}}\(=\)\dfrac{m}{n}\cdot\dfrac{\sqrt{a}}{\sqrt{b}}$；\(\boldsymbol{3}\).分母含有根号时\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)需要**分母有理化**；4.最终结果必须是最简二次根式：根号内无分母\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)分母不含根号\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)无开得尽方的因式。满分：100分时间：40分钟##一、选择题（每题4分\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)共20分）1.$\dfrac{\sqrt{15}}{\sqrt{\(\boldsymbol{3}\)}}$的结果是（）A.$\sqrt{5}$B.$5$C.$\sqrt{12}$D.$\(\boldsymbol{3}\)$2.计算$\sqrt{\dfrac{1}{2}}$化简结果是（）A.$\dfrac12$B.$\dfrac{\sqrt{2}}{2}$C.$\sqrt{2}$D.$2$\(\boldsymbol{3}\).下列计算正确的是（）A.$\dfrac{\sqrt{8}}{\sqrt{4}}\(=\)\sqrt{4}\(=\)2$B.$\dfrac{\sqrt{6}}{\sqrt{2}}\(=\)\(\boldsymbol{3}\)$C.$\sqrt{\dfrac{-4}{-9}}\(=\)\dfrac{\sqrt{-4}}{\sqrt{-9}}$D.$\dfrac{\sqrt{12}}{\sqrt{\(\boldsymbol{3}\)}}\(=\)2$4.等式$\sqrt{\dfrac{x}{x-2}}\(=\)\dfrac{\sqrt{x}}{\sqrt{x-2}}$成立的条件是（）A.$x\ge0$B.$\(x>2\)$C.$x\ge2$D.$0\lex<2$5.$\dfrac{4\sqrt{6}}{2\sqrt{2}}\(=\)$（）A.$2\sqrt{\(\boldsymbol{3}\)}$B.$2\sqrt{4}$C.$4\sqrt{\(\boldsymbol{3}\)}$D.$\sqrt{\(\boldsymbol{3}\)}$##二、填空题（每题4分\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)共20分）1.$\sqrt{\dfrac{25}{16}}\(=\)$________。2.$\dfrac{\sqrt{24}}{\sqrt{\(\boldsymbol{3}\)}}\(=\)$________。\(\boldsymbol{3}\).$\sqrt{\dfrac{\(\boldsymbol{3}\)}{4}}\(=\)$________。4.$\dfrac{\sqrt{18}}{\sqrt{2}}\(=\)$________。5.化简$\sqrt{\dfrac{2}{\(\boldsymbol{3}\)}}\(=\)$________。##三、基础解答题（每题8分\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)共\(\boldsymbol{3}\)2分）1.计算：（1）$\dfrac{\sqrt{20}}{\sqrt{5}}$（2）$\dfrac{\sqrt{48}}{\sqrt{6}}$2.计算：（1）$\dfrac{\(\boldsymbol{3}\)\sqrt{12}}{2\sqrt{\(\boldsymbol{3}\)}}$（2）$\dfrac{\sqrt{27}}{\sqrt{\(\boldsymbol{3}\)}}$\(\boldsymbol{3}\).化简：（1）$\sqrt{\dfrac{5}{12}}$（2）$\sqrt{\dfrac{2b}{\(\boldsymbol{3}\)a}}\quad(a>0,b\ge0)$4.分母有理化：（1）$\dfrac{1}{\sqrt{7}}$（2）$\dfrac{2}{\sqrt{6}}$##四、拓展提升题（每题14分\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)共28分）1.计算：$\sqrt{1\dfrac12}\div\sqrt{\dfrac16}$2.一个长方形面积是$\sqrt{48}\\text{cm}^2$\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)长是$\sqrt{8}\\text{cm}$\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)求长方形的宽。---#参考答案与解析##一、选择题1.A解析：$\dfrac{\sqrt{15}}{\sqrt{\(\boldsymbol{3}\)}}\(=\)\sqrt{\dfrac{15}{\(\boldsymbol{3}\)}}\(=\)\sqrt5$2.B解析：$\sqrt{\dfrac12}\(=\)\dfrac{1}{\sqrt2}\(=\)\dfrac{\sqrt2}{2}$\(\boldsymbol{3}\).D解析：A：$\sqrt2$；B：$\sqrt\(\boldsymbol{3}\)$；C：根号内不能为负数。4.B解析：$\begin{cases}x\ge0\\x-2>0\end{cases}\Rightarrow\(x>2\)$5.A解析：$\dfrac{4}{2}\cdot\dfrac{\sqrt6}{\sqrt2}\(=\)2\sqrt\(\boldsymbol{3}\)$##二、填空题1.$\boldsymbol{\dfrac54}$2.$\boldsymbol{2\sqrt2}$\(\boldsymbol{3}\).$\boldsymbol{\dfrac{\sqrt\(\boldsymbol{3}\)}{2}}$4.$\boldsymbol{\(\boldsymbol{3}\)}$5.$\boldsymbol{\dfrac{\sqrt6}{\(\boldsymbol{3}\)}}$##三、基础解答题1.（1）原式$\(=\)\sqrt{\dfrac{20}{5}}\(=\)\sqrt4\(=\)\boldsymbol{2}$（2）原式$\(=\)\sqrt{\dfrac{48}{6}}\(=\)\sqrt8\(=\)\boldsymbol{2\sqrt2}$2.（1）原式$\(=\)\dfrac\(\boldsymbol{3}\)2\sqrt{\dfrac{12}{\(\boldsymbol{3}\)}}\(=\)\dfrac\(\boldsymbol{3}\)2\sqrt4\(=\)\dfrac\(\boldsymbol{3}\)2\times2\(=\)\boldsymbol{\(\boldsymbol{3}\)}$（2）原式$\(=\)\sqrt{\dfrac{27}{\(\boldsymbol{3}\)}}\(=\)\sqrt9\(=\)\boldsymbol{\(\boldsymbol{3}\)}$\(\boldsymbol{3}\).（1）$\sqrt{\dfrac{5}{12}}\(=\)\sqrt{\dfrac{15}{\(\boldsymbol{3}\)6}}\(=\)\boldsymbol{\dfrac{\sqrt{15}}{6}}$（2）$\sqrt{\dfrac{2b}{\(\boldsymbol{3}\)a}}\(=\)\boldsymbol{\dfrac{\sqrt{6ab}}{\(\boldsymbol{3}\)a}}$4.（1）$\dfrac{1}{\sqrt7}\(=\)\boldsymbol{\dfrac{\sqrt7}{7}}$（2）$\dfrac{2}{\sqrt6}\(=\)\dfrac{2\sqrt6}{6}\(=\)\boldsymbol{\dfrac{\sqrt6}{\(\boldsymbol{3}\)}}$##四、拓展提升题1.解：$\begin{align}\text{原式}&\(=\)\sqrt{\dfrac\(\boldsymbol{3}\)2\div\dfrac16}\\&\(=\)\sqrt{\dfrac\(\boldsymbol{3}\)2\times6}\\&\(=\)\sqrt9\(=\)\boldsymbol{\(\boldsymbol{3}\)}\end{align}$2.解：宽$\(=\)$面积$\div$长$\begin{align}\text{宽}&\(=\)\sqrt{48}\div\sqrt{8}\\&\(=\)\sqrt{\dfrac{48}{8}}\(=\)\sqrt6\end{align}$答：长方形宽为$\boldsymbol{\sqrt6}\\text{cm}$##易错警示1.公式$\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}$要求$a\ge0\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)\boldsymbol{b>0}$\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)分母不能等于0；2.根号内是分数\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)不能直接把分子分母各自开方后保留分母根号；\(\boldsymbol{3}\).带系数的除法\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)系数单独相除\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)不要混入根号；4.结果一定要分母有理化\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b}}\(=\)\dfrac{\sqrt{a}}{\sqrt{b}}\quad(a\ge0\(a\ge0，\boldsymbol{b>0}\)b>0)}\)b>0)}\)分母不能带根号。下一节：**\(\boldsymbol{3}\).\(\boldsymbol{3}\).1二次根式的加减**\(\boldsymbol{\dfrac{\sqrt{a}}{\sqrt{b}}\(=\)\sqrt{\dfrac{a}{b}}\quad(a\ge0\(\boldsymbol{\sqrt{\dfrac{a}{b