标题

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2019年中考数学一轮复习数与式模块-整式专题复习讲义-中考数学复习资料

二.填空题(共6小题)

32.下面是按一定规律排列的代数式:\({a^2}\),\(3{a^4}\),\(5{a^6}\),\(7{a^8}\),\( \ldots \)则第8个代数式是 \(15{a^{16}}\) 

\({\color{red}{【解答】}}\)解:∵\( a^{2}\),\(3{a^4}\),\(5{a^6}\),\(7{a^8}\),\( \ldots \)

∴\( \)单项式的次数是连续的偶数,系数是连续的奇数,

∴\( \)第8个代数式是:\((2 \times 8 – 1){a^{2 \times 8}} = 15{a^{16}}\).

故答案为:\(15{a^{16}}\).

\({\color{red}{【总结】}}\)此题主要考查了单项式,正确得出单项式次数与系数的变化规律是解题关键.

33.已知\(m + n = 12\),\(m – n = 2\),则\({m^2} – {n^2} = \) 24 

\({\color{red}{【解答】}}\)解:∵\( m+n=12\),\(m – n = 2\),

∴\( {m^2} – {n^2} = (m + n)(m – n) = 2 \times 12 = 24\),

故答案为:24

\({\color{red}{【总结】}}\)此题考查平方差公式,关键是根据平方差公式的形式解答.

34.若\({2^x} = 5\),\({2^y} = 3\),则\({2^{2x + y}} = \) 75 

\({\color{red}{【解答】}}\)解:∵\( 2^{x}=5\),\({2^y} = 3\),

∴\( {2^{2x + y}} = {({2^x})^2} \times {2^y} = {5^2} \times 3 = 75\).

故答案为:75.

\({\color{red}{【总结】}}\)此题主要考查了同底数幂的乘法运算以及幂的乘方运算,正确掌握运算法则是解题关键.

35.已知\(m + n = mn\),则\((m – 1)(n – 1) = \) 1 .

\({\color{red}{【解答】}}\)解:\((m – 1)(n – 1) = mn – (m + n) + 1\),

∵\( m+n=mn\),

∴\( (m – 1)(n – 1) = mn – (m + n) + 1 = 1\),

故答案为1.

\({\color{red}{【总结】}}\)本题主要考查了整式的化简求值的知识,解答本题的关键是掌握多项式乘以多项式的运算法则,此题难度不大.

36.计算:\({(a + 1)^2} – {a^2} = \) \(2a + 1\) 

\({\color{red}{【解答】}}\)解:原式\( = {a^2} + 2a + 1 – {a^2} = 2a + 1\),

故答案为:\(2a + 1\)

\({\color{red}{【总结】}}\)此题考查了完全平方公式,熟练掌握完全平方公式是解本题的关键.

37.已知\({a^m} = 3\),\({a^n} = 2\),则\({a^{2m – n}}\)的值为 4.5 

\({\color{red}{【解答】}}\)解:∵\( a^{m}=3\),

∴\( {a^{2m}} = {3^2} = 9\),

∴\( {a^{2m – n}} = \frac{{{a^{2m}}}}{{{a^n}}} = \frac{9}{2} = 4.5\).

故答案为:4.5.

\({\color{red}{【总结】}}\)此题主要考查了同底数幂的除法法则,以及幂的乘方与积的乘方,同底数幂相除,底数不变,指数相减,要熟练掌握,解答此题的关键是要明确:①底数\(a \ne 0\),因为0不能做除数;②单独的一个字母,其指数是1,而不是0;③应用同底数幂除法的法则时,底数\(a\)可是单项式,也可以是多项式,但必须明确底数是什么,指数是什么.

三.解答题(共13小题)

38.先化简,再求值:\({(x – 1)^2} + x(3 – x)\),其中\(x =  – \frac{1}{2}\).

\({\color{red}{【解答】}}\)解:原式\( = {x^2} – 2x + 1 + 3x – {x^2} = x + 1\),

当\(x =  – \frac{1}{2}\)时,原式\( =  – \frac{1}{2} + 1 = \frac{1}{2}\).

\({\color{red}{【总结】}}\)此题主要考查了整式的混合运算\( –  – \)化简求值,关键是先按运算顺序把整式化简,再把对应字母的值代入求整式的值.

39.化简:\((y + 2)(y – 2) – (y – 1)(y + 5)\).

\({\color{red}{【解答】}}\)解:原式\( = {y^2} – 4 – {y^2} – 5y + y + 5 =  – 4y + 1\),

\({\color{red}{【总结】}}\)此题考查了整式的混合运算,熟练掌握运算法则是解本题的关键.

40.已知:\({x^2} – {y^2} = 12\),\(x + y = 3\),求\(2{x^2} – 2xy\)的值.

\({\color{red}{【解答】}}\)解:∵\( x^{2}-y^{2}=12\),

∴\( (x + y)(x – y) = 12\),

∵\( x+y=3\)①,

∴\( x – y = 4\)②,

①\( + \)②得,\(2x = 7\),

∴\( 2{x^2} – 2xy = 2x(x – y) = 7 \times 4 = 28\).

\({\color{red}{【总结】}}\)此题主要考查了平方差公式,二元一次方程的解法,求出\(x – y = 4\)是解本题的关键.

41.某同学化简\(a(a + 2b) – (a + b)(a – b)\)出现了错误,解答过程如下:

原式\( = {a^2} + 2ab – ({a^2} – {b^2})\) (第一步)

\( = {a^2} + 2ab – {a^2} – {b^2}\)(第二步)

\( = 2ab – {b^2}\) (第三步)

(1)该同学解答过程从第 二 步开始出错,错误原因是  

(2)写出此题正确的解答过程.

\({\color{red}{【解答】}}\)解:(1)该同学解答过程从第 二步开始出错,错误原因是 去括号时没有变号;

故答案是:二;去括号时没有变号;

(2)原式\( = {a^2} + 2ab – ({a^2} – {b^2})\)

\( = {a^2} + 2ab – {a^2} + {b^2}\)

\( = 2ab + {b^2}\).

\({\color{red}{【总结】}}\)考查了平方差公式和实数的运算,去括号规律:①\(a + (b + c) = a + b + c\),括号前是“\( + \)”号,去括号时连同它前面的“\( + \)”号一起去掉,括号内各项不变号;②\(a – (b – c) = a – b + c\),括号前是“\( – \)”号,去括号时连同它前面的“\( – \)”号一起去掉,括号内各项都要变号.

42.先化简,再求值:\(a(a + 2b) – {(a + 1)^2} + 2a\),其中\(a = \sqrt 2  + 1,b = \sqrt 2  – 1\).

\({\color{red}{【解答】}}\)解:原式\( = {a^2} + 2ab – ({a^2} + 2a + 1) + 2a\)

\( = {a^2} + 2ab – {a^2} – 2a – 1 + 2a\)

\( = 2ab – 1\),

当\(a = \sqrt 2  + 1,b = \sqrt 2  – 1\)时,

原式\( = 2(\sqrt 2  + 1)(\sqrt 2  – 1) – 1\)

\( = 2 – 1\)

\( = 1\).

\({\color{red}{【总结】}}\)本题考查了整式的混合运算\(-\)化简求值,能正确根据整式的运算法则进行化简是解此题的关键.

43.先化简,再求值:\((a – 2b)(a + 2b) – {(a – 2b)^2} + 8{b^2}\),其中\(a =  – 2\),\(b = \frac{1}{2}\).

\({\color{red}{【解答】}}\)解:原式\( = {a^2} – 4{b^2} – {a^2} + 4ab – 4{b^2} + 8{b^2} = 4ab\),

当\(a =  – 2\),\(b = \frac{1}{2}\)时,原式\( =  – 4\).

\({\color{red}{【总结】}}\)此题考查了整式的混合运算\( – \)化简求值,熟练掌握运算法则是解本题的关键.

44.先化简,再求值:\((x + 2)(x – 2) + x(1 – x)\),其中\(x =  – 1\).

\({\color{red}{【解答】}}\)解:原式\( = {x^2} – 4 + x – {x^2} = x – 4\),

当\(x =  – 1\)时,原式\( =  – 5\).

\({\color{red}{【总结】}}\)此题考查了整式的混合运算\( – \)化简求值,熟练掌握运算法则是解本题的关键.

45.先化简,再求值:\({(a + b)^2} + b(a – b) – 4ab\),其中\(a = 2\),\(b =  – \frac{1}{2}\).

\({\color{red}{【解答】}}\)解:原式\( = {a^2} + 2ab + {b^2} + ab – {b^2} – 4ab = {a^2} – ab\),

当\(a = 2\),\(b =  – \frac{1}{2}\)时,原式\( = 4 + 1 = 5\).

\({\color{red}{【总结】}}\)此题主要考查了整式的混合运算\( –  – \)化简求值,关键是先按运算顺序把整式化简,再把对应字母的值代入求整式的值.

46.有一张边长为\(a\)厘米的正方形桌面,因为实际需要,需将正方形边长增加\(b\)厘米,木工师傅设计了如图所示的三种方案:

小明发现这三种方案都能验证公式:\({a^2} + 2ab + {b^2} = {(a + b)^2}\),

对于方案一,小明是这样验证的:

\({a^2} + ab + ab + {b^2} = {a^2} + 2ab + {b^2} = {(a + b)^2}\)

请你根据方案二、方案三,写出公式的验证过程.

方案二:

方案三:

\({\color{red}{【解答】}}\)解:由题意可得,

方案二:\({a^2} + ab + (a + b)b = {a^2} + ab + ab + {b^2} = {a^2} + 2ab + {b^2} = {(a + b)^2}\),

方案三:\({a^2} + \frac{{[a + (a + b)]b}}{2} + \frac{{[a + (a + b)]b}}{2} = {a^2} + ab + \frac{1}{2}{b^2} + ab + \frac{1}{2}{b^2} = {a^2} + 2ab + {b^2} = {(a + b)^2}\).

\({\color{red}{【总结】}}\)本题考查完全平方公式的几何背景,解答本题的关键是明确题意,写出相应的推导过程.

47.先化简, 再求值:\((2m + 1)(2m – 1) – {(m – 1)^2} + {(2m)^3} \div ( – 8m)\),其中\(m\)是方程\({x^2} + x – 2 = 0\)的根

\({\color{red}{【解答】}}\)解: 原式\( = 4{m^2} – 1 – ({m^2} – 2m + 1) + 8{m^3} \div ( – 8m)\)

\( = 4{m^2} – 1 – {m^2} + 2m – 1 – {m^2}\)

\( = 2{m^2} + 2m – 2\)

\( = 2({m^2} + m – 1)\),

∵\( m\)是方程\({x^2} + x – 2 = 0\)的根,

∴\( {m^2} + m – 2 = 0\),即\({m^2} + m = 2\),

则原式\( = 2 \times (2 – 1) = 2\).

\({\color{red}{【总结】}}\)本题主要考查整式的化简求值, 解题的关键是掌握平方差公式和完全平方公式、 整式的混合运算顺序和运算法则、 方程的解的定义 .

48.先化简,再求值:\((x + 1)(x – 1) + {(2x – 1)^2} – 2x(2x – 1)\),其中\(x = \sqrt 2  + 1\).

\({\color{red}{【解答】}}\)解:原式\( = {x^2} – 1 + 4{x^2} – 4x + 1 – 4{x^2} + 2x\)

\( = {x^2} – 2x\),

把\(x = \sqrt 2  + 1\)代入,得:

原式\( = {(\sqrt 2  + 1)^2} – 2(\sqrt 2  + 1)\)

\( = 3 + 2\sqrt 2  – 2\sqrt 2  – 2\)

\( = 1\).

\({\color{red}{【总结】}}\)本题考查了整式的混合运算及化简求值,做好本题要熟练掌握多项式乘以多项式的法则和整式乘法公式,此类题的思路为:先按运算顺序把整式化简,再把对应字母的值代入求整式的值.

49.阅读以下材料:

对数的创始人是苏格兰数学家纳皮尔\((J\).\(Nplcr\),\(1550 – 1617\)年),纳皮尔发明

对数是在指数书写方式之前,直到 18 世纪瑞士数学家欧拉\((Evlcr\),\(1707 – 1783\)

年)才发现指数与对数之间的联系 .

对数的定义: 一般地, 若\({a^x} = N(a > 0,a \ne 1)\),那么\(x\)叫做以\(a\)为底\(N\)的对数,

记作:\(x = {\log _a}N\). 比如指数式\({2^4} = 16\)可以转化为\(4 = {\log _2}16\),对数式\(2 = {\log _5}25\)

可以转化为\({5^2} = 25\).

我们根据对数的定义可得到对数的一个性质:

\(\log _{a}(M· N)=\log _{a}M+\log _{a}N(a>0\),\(a \ne 1\),\(M > 0\),\(N > 0)\);理由如下:

设\({\log _a}M = m\),\({\log _a}N = n\),则\(M = {a^m}\),\(N = {a^n}\)

∴\( M· N=a^{m}· a^{n}=a^{m+n}\),由对数的定义得\(m+n=\log _{a}(M· N)\)

又∵\( m+n=\log _{a}M+\log _{a}N\)

∴\( \log _{a}(M· N)=\log _{a}M+\log _{a}N\)

解决以下问题:

(1) 将指数\({4^3} = 64\)转化为对数式 \(3 = {\log _4}64\) ;

(2) 证明\({\log _a}\frac{M}{N} = {\log _a}M – {\log _a}N(a > 0\),\(a \ne 1\),\(M > 0\),\(N > 0)\)

(3) 拓展运用: 计算\({\log _3}2 + {\log _3}6 – {\log _3}4 = \)  

\({\color{red}{【解答】}}\)解: (1) 由题意可得, 指数式\({4^3} = 64\)写成对数式为:\(3 = {\log _4}64\),

故答案为:\(3 = {\log _4}64\);

(2) 设\({\log _a}M = m\),\({\log _a}N = n\),则\(M = {a^m}\),\(N = {a^n}\),

∴\( \)\(\frac{M}{N} = \frac{{{a^m}}}{{{a^n}}} = {a^{m – n}}\),由对数的定义得\(m – n = {\log _a}\frac{M}{N}\),

又∵\( m-n=\log _{a}M-\log _{a}N\),

∴\( {\log _a}\frac{M}{N} = {\log _a}M – {\log _a}N(a > 0\),\(a \ne 1\),\(M > 0\),\(N > 0)\);

(3)\({\log _3}2 + {\log _3}6 – {\log _3}4\),

\( = {\log _3}(2 \times 6 \div 4)\),

\( = {\log _3}3\),

\( = 1\),

故答案为: 1 .

\({\color{red}{【总结】}}\)本题考查整式的混合运算、 对数与指数之间的关系与相互转化的关系, 解题的关键是明确新定义, 明白指数与对数之间的关系与相互转化关系 .

50.嘉淇准备完成题目:

发现系数“□”印刷不清楚.

(1)他把“□”猜成3,请你化简:\((3{x^2} + 6x + 8) – (6x + 5{x^2} + 2)\);

(2)他妈妈说:“你猜错了,我看到该题标准答案的结果是常数.”通过计算说明原题中“□”是几?

\({\color{red}{【解答】}}\)解:(1)\((3{x^2} + 6x + 8) – (6x + 5{x^2} + 2)\)

\( = 3{x^2} + 6x + 8 – 6x – 5{x^2} – 2\)

\( =  – 2{x^2} + 6\);

(2)设“□”是\(a\),

则原式\( = (a{x^2} + 6x + 8) – (6x + 5{x^2} + 2)\)

\( = a{x^2} + 6x + 8 – 6x – 5{x^2} – 2\)

\( = (a – 5){x^2} + 6\),

∵\(\)标准答案的结果是常数,

∴\( a – 5 = 0\),

解得:\(a = 5\).

\({\color{red}{【总结】}}\)本题主要考查整式的加减,解题的关键是掌握去括号、合并同类项法则.

2019.3.23



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