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2019年中考复习数与式版块—二次根式专题复习讲义_2019年中考数学复习资料

2018年全国各省市二次根式计算中考真题汇编知识点分析
知识点题目量占比
最简二次根式24.00%
分式有意义的条件12.00%
二次根式的加减法48.00%
同类二次根式12.00%
合并同类项48.00%
同底数幂的乘法12.00%
二次根式的混合运算612.00%
二次根式的性质与化简24.00%
二次根式有意义的条件1326.00%
二次根式的乘除法48.00%
同底数幂的除法12.00%
实数与数轴12.00%
二次根式的应用24.00%
分母有理化12.00%
负整数指数幂24.00%
规律型:数字的变化类12.00%
零指数幂24.00%
实数的运算24.00%

2018年全国各省市中考真题汇编—二次根式专题复习讲义

一.选择题(共21小题)

1.(2018•兰州)下列二次根式中,是最简二次根式的是\((\)  \()\)

A.\(\sqrt {18} \)

B.\(\sqrt {13} \)

C.\(\sqrt {27} \)

D.\(\sqrt {12} \)

2.(2018•绥化)若\(y = \frac{{\sqrt {1 – 2x} }}{x}\)有意义,则\(x\)的取值范围是\((\)  \()\)

A.\(x\leq\frac{1}{2}\)且\(x \ne 0\)

B.\(x \ne \frac{1}{2}\)

C.\(x\leq \frac{1}{2}\)

D.\(x \ne 0\)

3.(2018•上海)下列计算\(\sqrt {18}  – \sqrt 2 \)的结果是\((\)  \()\)

A.4

B.3

C.\(2\sqrt 2 \)

D.\(\sqrt 2 \)

4.(2018•曲靖)下列二次根式中能与\(2\sqrt 3 \)合并的是\((\)  \()\)

A.\(\sqrt 8 \)

B.\(\sqrt {\frac{1}{3}} \)

C.\(\sqrt {18} \)

D.\(\sqrt 9 \)

5.(2018•张家界)下列运算正确的是\((\)  \()\)

A.\({a^2} + a = 2{a^3}\)

B.\(\sqrt {{a^2}}  = a\)

C.\({(a + 1)^2} = {a^2} + 1\)

D.\({({a^3})^2} = {a^6}\)

6.(2018•郴州)下列运算正确的是\((\)  \()\)

A.\(a^{3}· a^{2}=a^{6}\)

B.\({a^{ – 2}} =  – \frac{1}{{{a^2}}}\)

C.\(3\sqrt 3  – 2\sqrt 3  = \sqrt 3 \)

D.\((a + 2)(a – 2) = {a^2} + 4\)

7.(2018•孝感)下列计算正确的是\((\)  \()\)

A.\({a^{ – 2}} \div {a^5} = \frac{1}{{{a^7}}}\)

B.\({(a + b)^2} = {a^2} + {b^2}\)

C.\(2 + \sqrt 2  = 2\sqrt 2 \)

D.\({({a^3})^2} = {a^5}\)

8.(2018•泰州)下列运算正确的是\((\)  \()\)

A .\(\sqrt 2  + \sqrt 3  = \sqrt 5 \)

B .\(\sqrt {18}  = 2\sqrt 3 \)

C .\(\sqrt{2}· \sqrt{3}=\sqrt{5}\)

D .\(\sqrt 2  \div \sqrt {\frac{1}{2}}  = 2\)

9.(2018•无锡)下列等式正确的是\((\)  \()\)

A.\({(\sqrt 3 )^2} = 3\)

B.\(\sqrt {{{( – 3)}^2}}  =  – 3\)

C.\(\sqrt {{3^3}}  = 3\)

D.\({( – \sqrt 3 )^2} =  – 3\)

10.(2018•扬州)使\(\sqrt {x – 3} \)有意义的\(x\)的取值范围是\((\)  \()\)

A.\(x > 3\)

B.\(x < 3\)

C.\(x\geq 3\)

D.\(x \ne 3\)

11.(2018•长沙)下列计算正确的是\((\)  \()\)

A.\({a^2} + {a^3} = {a^5}\)

B.\(3\sqrt 2  – 2\sqrt 2  = 1\)

C.\({({x^2})^3} = {x^5}\)

D.\({m^5} \div {m^3} = {m^2}\)

12.(2018•自贡)下列计算正确的是\((\)  \()\)

A.\({(a – b)^2} = {a^2} – {b^2}\)

B.\(x + 2y = 3xy\)

C.\(\sqrt {18}  – 3\sqrt 2  = 0\)

D.\({( – {a^3})^2} =  – {a^6}\)

13.(2018•达州)二次根式\(\sqrt {2x + 4} \)中的\(x\)的取值范围是\((\)  \()\)

A.\(x <  – 2\)

B.\(x\leq-2\)

C.\(x >  – 2\)

D.\(x\geq -2\)

14.(2018•绵阳)等式\(\frac{{\sqrt {x – 3} }}{{\sqrt {x + 1} }} = \sqrt {\frac{{x – 3}}{{x + 1}}} \)成立的\(x\)的取值范围在数轴上可表示为\((\)  \()\)

15.(2018•怀化)使\(\sqrt {x – 3} \)有意义的\(x\)的取值范围是\((\)  \()\)

A.\(x\leq3\)

B.\(x < 3\)

C.\(x\geq 3\)

D.\(x > 3\)

16.(2018•抚顺)二次根式\(\sqrt {1 – x} \)在实数范围内有意义,则\(x\)的取值范围是\((\)  \()\)

A.\(x\geq 1\)

B.\(x\leq1\)

C.\(x > 1\)

D.\(x < 1\)

17.(2018•南通)若代数式\(\sqrt {x – 1} \)在实数范围内有意义,则\(x\)的取值范围是\((\)  \()\)

A.\(x < 1\)

B.\(x\leq 1\)

C.\(x > 1\)

D.\(x\geq 1\)

18.(2018•临安区)下列各式计算正确的是\((\)  \()\)

A.\({a^{12}} \div {a^6} = {a^2}\)

B.\({(x + y)^2} = {x^2} + {y^2}\)

C.\(\frac{{x – 2}}{{4 – {x^2}}} = \frac{1}{{2 + x}}\)

D.\(\sqrt {\frac{3}{5}}  \div \sqrt 5  = \frac{{\sqrt 3 }}{5}\)

19.(2018•德阳)下列计算或运算中,正确的是\((\)  \()\)

A.\(2\sqrt {\frac{a}{2}}  = \sqrt a \)

B.\(\sqrt {18}  – \sqrt 8  = \sqrt 2 \)

C.\(6\sqrt {15}  \div 2\sqrt 3  = 3\sqrt {45} \)

D.\( – 3\sqrt 3  = \sqrt {27} \)

20.(2018•日照)若式子\(\frac{{\sqrt {m + 2} }}{{{{(m – 1)}^2}}}\)有意义,则实数\(m\)的取值范围是\((\)  \()\)

A.\(m >  – 2\)

B.\(m >  – 2\)且\(m \ne 1\)

C.\(m\geq -2\)

D.\(m\geq -2\)且\(m \ne 1\)

21.(2018•聊城)下列计算正确的是\((\)  \()\)

A.\(3\sqrt {10}  – 2\sqrt 5  = \sqrt 5 \)

B.\(\sqrt{\frac{7}{11}}· (\sqrt{\frac{11}{7}}\div \sqrt{\frac{1}{11}})=\sqrt{11}\)

C.\((\sqrt {75}  – \sqrt {15} ) \div \sqrt 3  = 2\sqrt 5 \)

D.\(\frac{1}{3}\sqrt {18}  – 3\sqrt {\frac{8}{9}}  = \sqrt 2 \)

二.填空题(共23小题)

22.(2018•连云港)使\(\sqrt {x – 2} \)有意义的\(x\)的取值范围是  

23.(2018•镇江)计算:\(\sqrt {\frac{1}{2}}  \times \sqrt 8  = \)  

24.(2018•广西)要使二次根式\(\sqrt {x – 5} \)在实数范围内有意义,则实数\(x\)的取值范围是  

25.(2018•广州)如图,数轴上点\(A\)表示的数为\(a\),化简:\(a + \sqrt {{a^2} – 4a + 4}  = \)  

26.(2018•巴中)已知\(|\sin A – \frac{1}{2}| + \sqrt {{{(\sqrt 3  – \tan B)}^2}}  = 0\),那么\(\angle A + \angle B = \)  

27.(2018•贺州)要使二次根式\(\sqrt {x – 3} \)有意义,则\(x\)的取值范围是  

28.(2018•盘锦)若式子\(\sqrt {2 – x}  + \sqrt {x – 1} \)有意义,则\(x\)的取值范围是  

29.(2018•烟台)\(\sqrt {12} \)与最简二次根式\(5\sqrt {a + 1} \)是同类二次根式,则\(a = \)  

30.(2018•哈尔滨)计算\(6\sqrt 5  – 10\sqrt {\frac{1}{5}} \)的结果是  

31.(2018•盘锦)计算:\(\sqrt {27}  – \sqrt {12}  = \)  

32.(2018•益阳)计算:\(\sqrt {12}  \times \sqrt 3  = \)   

33.(2018•郴州)计算:\({( – \sqrt 3 )^2} = \)  

34.(2018•南京)若式子\(\sqrt {x – 2} \)在实数范围内有意义, 则\(x\)的取值范围是  

35.(2018•莱芜)如图,正三角形和矩形具有一条公共边,矩形内有一个正方形,其四个顶点都在矩形的边上,正三角形和正方形的面积分别是\(2\sqrt 3 \)和2,则图中阴影部分的面积是  

36.(2018•毕节市)观察下列运算过程:

\(\begin{array}{*{20}{l}}{\frac{1}{{1 + \sqrt 2 }} = \frac{1}{{\sqrt 2  + 1}} = \frac{{\sqrt 2  – 1}}{{(\sqrt 2  + 1)(\sqrt 2  – 1)}} = \frac{{\sqrt 2  – 1}}{{{{(\sqrt 2 )}^2} – {1^2}}} = \sqrt 2  – 1}\\{\frac{1}{{\sqrt 2  + \sqrt 3 }} = \frac{1}{{\sqrt 3  + \sqrt 2 }} = \frac{{\sqrt 3  – \sqrt 2 }}{{(\sqrt 3  + \sqrt 2 )(\sqrt 3  – \sqrt 2 )}} = \frac{{\sqrt 3  – \sqrt 2 }}{{{{(\sqrt 3 )}^2} – {{(\sqrt 2 )}^2}}} = \sqrt 3  – \sqrt 2 }\\{ \ldots  \ldots }\end{array}\)

请运用上面的运算方法计算:

\(\frac{1}{{1 + \sqrt 3 }} + \frac{1}{{\sqrt 3  + \sqrt 5 }} + \frac{1}{{\sqrt 5  + \sqrt 7 }} +  \ldots  + \frac{1}{{\sqrt {2015}  + \sqrt {2017} }} + \frac{1}{{\sqrt {2017}  + \sqrt {2019} }} = \)  

37.(2018•广安)要使\(\sqrt {x + 1} \)有意义,则实数\(x\)的取值范围是  

38.(2018•山西)计算:\((3\sqrt 2  + 1)(3\sqrt 2  – 1) = \)  

39.(2018•湖北)计算:\(\frac{3}{{\sqrt 3 }} + |\sqrt 3  – 2| – {(\frac{1}{2})^{ – 1}} = \)  

40.(2018•南京)计算\(\sqrt 3  \times \sqrt 6  – \sqrt 8 \)的结果是  

41.(2018•湖州)二次根式\(\sqrt {x – 3} \)中字母\(x\)的取值范围是  

42.(2018•天津)计算\((\sqrt 6  + \sqrt 3 )(\sqrt 6  – \sqrt 3 )\)的结果等于  

43.(2018•滨州)观察下列各式:

\(\sqrt {1 + \frac{1}{{{1^2}}} + \frac{1}{{{2^2}}}}  = 1 + \frac{1}{{1 \times 2}}\),

\(\sqrt {1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}}}  = 1 + \frac{1}{{2 \times 3}}\),

\(\sqrt {1 + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}}}  = 1 + \frac{1}{{3 \times 4}}\),

\( \ldots  \ldots \)

请利用你所发现的规律,

计算\(\sqrt {1 + \frac{1}{{{1^2}}} + \frac{1}{{{2^2}}}}  + \sqrt {1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}}}  + \sqrt {1 + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}}}  +  \ldots  + \sqrt {1 + \frac{1}{{{9^2}}} + \frac{1}{{{{10}^2}}}} \),其结果为  

44.(2018•枣庄)我国南宋著名数学家秦九韶在他的著作《数书九章》一书中,给出了著名的秦九韶公式,也叫三斜求积公式,即如果一个三角形的三边长分别为\(a\),\(b\),\(c\),则该三角形的面积为\(S = \sqrt {\frac{1}{4}[{a^2}{b^2} – {{(\frac{{{a^2} + {b^2} – {c^2}}}{2})}^2}]} \).现已知\(\Delta ABC\)的三边长分别为1,2,\(\sqrt 5 \),则\(\Delta ABC\)的面积为  

三.解答题(共6小题)

45.(2018•柳州)计算:\(2\sqrt 4  + 3\).

46.(2018•大连)计算:\({(\sqrt 3  + 2)^2} – \sqrt {48}  + {2^{ – 2}}\).

47.(2018•德阳)计算:\(\sqrt {{{( – 3)}^2}}  + {(\frac{1}{2})^{ – 3}} – {(3\sqrt 2 )^0} – 4\cos 30^\circ  + \frac{6}{{\sqrt 3 }}\).

48.(2018•兰州)计算:\({( – \frac{1}{2})^{ – 2}} + {(\pi  – 3)^0} + |1 – \sqrt 2 | + \tan 45^\circ \).

49.(2018•梧州)计算:\(\sqrt 9  – {2^5} \div {2^3} + | – 1| \times 5 – {(\pi  – 3.14)^0}\).

50.(2018•陕西)计算:\(( – \sqrt 3 ) \times ( – \sqrt 6 ) + |\sqrt 2  – 1| + {(5 – 2\pi )^0}\).

2019.3.22



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2019年中考复习:2018年全国各省市中考真题汇编—分式(一)

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