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中考专题18 新定义与阅读理解题(解析版)

中考专题18 新定义与阅读理解题(解析版)
中考专题18 新定义与阅读理解题(解析版)

专题18 新定义与阅读理解题

1.(2019?湘西州)阅读材料:设a r

=(x 1,y 1),b r =(x 2,y 2),如果

a r ∥

b r ,则x 1?y 2=x 2?y 1,根据该材料填空,已知a r

=(4,3),b r =(8,m ),且a r ∥b r ,则

m =__________.

【答案】6

【解析】∵a r =(4,3),b r =(8,m ),且a r ∥b r ,∴4m =3×8,∴m =6;

故答案为:6.

【名师点睛】本题考查新定义,点的坐标;理解阅读材料的内容,转化为所学知识求解是关键.

2.(2019?白银)定义:等腰三角形的顶角与其一个底角的度数的比

值k 称为这个等腰三角形的“特征值”.若等腰△ABC 中,∠A =80°,则它的特征值k =__________. 【答案】8

5或1

4

【解析】①当∠A 为顶角时,等腰三角形两底角的度数为:2

18080?-?

=50°, ∴特征值

k =808505?=?;

②当∠A 为底角时,顶角的度数为:180°–80°–80°=20°, ∴特征值

k =208014?=?;

综上所述,特征值k

为85或14;

故答案为85或1

4.

【名师点睛】本题主要考查等腰三角形的性质,熟记等腰三角形的性质是解题的关键,要注意到本题中,已知∠A 的底数,要进行判断

是底角或顶角,以免造成答案的遗漏.

3.(2019?河北)如图,约定:上方相邻两数之和等于这两数下方箭头共同指向的数.

示例:即4+3=7.

则(1)用含x的式子表示m=__________;

(2)当y=–2时,n的值为__________.

【答案】(1)3x;(2)1.

【解析】(1)根据约定的方法可得:m=x+2x=3x;故答案为:3x;(2)根据约定的方法即可得x+2x+2x+3=m+n=y.

当y=–2时,5x+3=–2.

解得x=–1.

∴n=2x+3=–2+3=1.

故答案为:1.

【名师点睛】本题考查了列代数式和代数式求值,解题的关键是掌握列代数式的约定方法.

4.(2019?枣庄)对于实数a、b,定义关于“?”的一种运算:a?b=2a+b,例如3?4=2×3+4=10.

(1)求4?(–3)的值;

(2)若x ?(–y )=2,(2y )?x =–1,求x +y 的值.

【答案】(1)5;(2)1

3.

【解析】(1)根据题中的新定义得:原式=8–3=5;

(2)根据题中的新定义化简得:2241x y x y -=?+=-??

②,

①+②得:3x +3y =1,则

x +y =13.

【名师点睛】此题考查了解二元一次方程组,以及实数的运算,熟练掌握运算法则是解本题的关键. 5.(2019?济宁)阅读下面的材料:

如果函数y =f (x )满足:对于自变量x 的取值范围内的任意x 1,x 2, (1)若x 1

x (x >0)是减函数. 证明:设0

f (x 1)–f (x 2)=()

2121121212

66666x x x x x x x x x x ---==.

∵0

∴()

2112

6x x x x ->0.即

f (x 1)–f (x 2)>0.

∴f (x 1)>f (x 2),∴函数

f (x )═6

x

(x >0)是减函数.

根据以上材料,解答下面的问题: 已知函数

f (x )=2

1

x +x (x <0),

f (–1)=21

(1)-+(–1)=0,f (–2)=2

1

(2)-+(–2)=–7

4

(1)计算:f (–3)=__________,f (–4)=__________; (2)猜想:函数f (x )=2

1

x +x (x <0)是__________函数(填“增”

或“减”);

(3)请仿照例题证明你的猜想.

【答案】(1)–269,–63

16;(2)增;(3)见解析.

【解析】(1)∵f (x )=21

x +x (x <0),

∴f (–3)=2

1

(3)-–3=–26

9,f (–4)=2

1(4)-–4=–63

16,

故答案为:–269,–63

16;

(2)∵–4<–3,f (–4)>f (–3), ∴函数f (x )=21

x +x (x <0)是增函数, 故答案为:增; (3)设x 1

∵f (x 1)–f (x 2)=122212

11

x x x x +--=(x 1–x 2)(1–1222

12x x x x +)

∵x 1

∴f (x 1)–f (x 2)<0,∴f (x 1)

f (x )=2

1

x +x (x <0)是增函数.

【名师点睛】本题考查反比例函数图象上的坐标特征、反比例函数的性质,解答本题的关键是明确题意,找出所求问题需要的条件,利用反比例函数的性质解答.

6.(2019?随州)若一个两位数十位、个位上的数字分别为m,n,我们可将这个两位数记为mn,易知mn=10m+n;同理,一个三位数、四位数等均可以用此记法,如abc=100a+10b+c.

【基础训练】

(1)解方程填空:

①若2x+3x=45,则x=__________;

②若7y–8y=26,则y=__________;

③若93t+58t=131t,则t=__________;

【能力提升】

(2)交换任意一个两位数mn的个位数字与十位数字,可得到一个新数nm,则mn+nm一定能被__________整除,mn–nm一定能被

__________整除,mn?nm–mn一定能被__________整除;(请从大于5的整数中选择合适的数填空)

【探索发现】

(3)北京时间2019年4月10日21时,人类拍摄的首张黑洞照片问世,黑洞是一种引力极大的天体,连光都逃脱不了它的束缚.数学中也存在有趣的黑洞现象:任选一个三位数,要求个、十、百位的数字各不相同,把这个三位数的三个数字按大小重新排列,得出一个最大的数和一个最小的数,用得出的最大的数减去最小的数得到一个新数(例如若选的数为325,则用532–235=297),再将这个新数按上述方式重新排列,再相减,像这样运算若干次后一定会得到同一个重复出现的数,这个数称为“卡普雷卡尔黑洞数”.

①该“卡普雷卡尔黑洞数”为__________;

②设任选的三位数为abc(不妨设a>b>c),试说明其均可产生该黑洞数.

【答案】(1)①2.②4.③7.(2)11;9;10.

【解析】(1)①∵mn=10m+n,

∴若2x+3x=45,则10×2+x+10x+3=45,

∴x=2,

故答案为:2.

②若7y–8y=26,则10×7+y–(10y+8)=26,

解得y=4,

故答案为:4.

③由abc=100a+10b+c,及四位数的类似公式得

若93t+58t=131t,则100t+10×9+3+100×5+10t+8=1000×1+100×

3+10t+1,

∴100t=700,

∴t=7,

故答案为:7.

(2)∵mn+nm=10m+n+10n+m=11m+11n=11(m+n),

∴则mn+nm一定能被11整除,

∵mn–nm=10m+n–(10n+m)=9m–9n=9(m–n),

∴mn–nm一定能被9整除.

∵mn?nm–mn=(10m+n)(10n+m)–mn=100mn+10m2+10n2+mn–mn=10(10mn+m2+n2)

∴mn?nm–mn一定能被10整除.

故答案为:11;9;10.

(3)①若选的数为325,则用532–235=297,以下按照上述规则继续计算,

972–279=693,

963–369=594,

954–459=495,

954–459=495,…

故答案为:495.

②当任选的三位数为abc时,第一次运算后得:100a+10b+c–

(100c+10b+a)=99(a–c),

结果为99的倍数,由于a>b>c,故a≥b+1≥c+2,

∴a–c≥2,又9≥a>c≥0,

∴a–c≤9,

∴a–c=2,3,4,5,6,7,8,9,

∴第一次运算后可能得到:198,297,396,495,594,693,792,891,

再让这些数字经过运算,分别可以得到:

981–189=792,972–279=693,963–369=594,954–459–495,954–459=495…,

故都可以得到该黑洞数495.

【名师点睛】本题是较为复杂的新定义试题,题目设置的问题较多,但解答方法大同小异,总体中等难度略大.

7.(2019?自贡)阅读下列材料:小明为了计算1+2+22+…+22017+22018

的值,采用以下方法: 设S =1+2+22+…+22017+22018①, 则2S =2+22+…+22018+22019②, ②–①得2S –S =S =22019–1, ∴S =1+2+22+…+22017+22018=22019–1. 请仿照小明的方法解决以下问题: (1)1+2+22+…+29=__________; (2)3+32+…+310=__________;

(3)求1+a +a 2+…+a n 的和(a >0,n 是正整数),请写出计算过程. 【答案】(1)2

10

–1;(2)1131

2-;(3)a =1

时,S =n +1;a ≠1时,

S =111

n a a +--. 【解析】(1)设S =1+2+22+…+29①, 则2S =2+22+…+210②, ②–①得2S –S =S =210–1, ∴S =1+2+22+…+29=210–1; 故答案为:210–1;

(2)设S =3+3+32+33+34+…+310①,

则3S =32+33+34+35+…+311②, ②–①得2S =311–1, 所以S =11312

-, 即

3+32+33+34+

+310=

11

31

2

-;

故答案为:1131

2-;

(3)设S =1+a +a 2+a 3+a 4+…+a n ①, 则aS =a +a 2+a 3+a 4+…+a n +a n +1②, ②–①得:(a –1)S =a n +1–1,

a =1时,不能直接除以a –1,此时原式等于n +1; a ≠1时,a –1才能做分母,所以S =111

n a a +--, 即1+a +a 2

+a 3

+a 4

+…+a

n

=11

1

n a a +--. 【名师点睛】根据题目给出的信息,提炼解题方法.认真观察、仔细思考,善用联想,利用类比的方法是解决这类问题的方法. 8.(2019·江西)特例感知

(1)如图1,对于抛物线211y x x =--+,2221y x x =--+,2331y x x =--+,下列结论正确的序号是_________; ①抛物线1y ,2y ,3y 都经过点(0,1)C ;

②抛物线2y ,3y 的对称轴由抛物线1y 的对称轴依次向左平移12

个单位

得到;

③抛物线1y ,2y ,3y 与直线1y =的交点中,相邻两点之间的距离相等. 形成概念

(2)把满足21n y x nx =--+(n 为正整数)的抛物线称为“系列平移抛物线”. 知识应用

在(2)中,如图2.

①“系列平移抛物线”的顶点依次为1P ,2P ,3P ,…,n P ,用含n 的代数式表示顶点n P 的坐标,并写出该顶点纵坐标y 与横坐标x 之间的关系式;

②“系列平移抛物线”存在“系列整数点(横、纵坐标均为整数的点)

”:1C ,2C ,3C ,…,n C ,其横坐标分别为:1k --,2k --,3k --,…,k n

--(k 为正整数),判断相邻两点之间的距离是否都相等,若相等,直接写出相邻两点之间的距离;若不相等,说明理由.

③在②中,直线1y =分别交“系列平移抛物线”于点1A ,2A ,3A ,…,n A ,连接n n C A ,11n n C A --,判断n n C A ,11n n C A --是否平行?并说明理由.

【答案】(1)①②③ (2)①

2,124n n n P ??

-+ ???

,2

1

y x

=+.

2

1k +

③不平行,直线n

n

C A 的斜率(比例系数)为k n +,与n 取值有关(若两直线平行,则斜率会相等). 【解析】(1)①当x =0,1

231y

y y ===,所以正确;

②1

2

3

,,y y y 的对称轴分别是直线1

12

x

=-

,2

1x

=-,332

x =-

,所以正确;

③1

2

3

,,y y y 与1y =交点(除了点C )横坐标分别为–1,–2,–3,所以距离为1,都相等,正确.

(2)①2

22

4124n n n y x nx x +?

?=--+=-++

??

?,所以顶点

24,2

4n n n P ??

+- ?

??,

令顶点n P 横坐标

2

n

x =-

,纵坐标24

4n y +=,2

2241142n n y x +??==-+=+ ???

, 即:n

P 顶点满足关系式2

1

y x

=+.

②相邻两点之间的距离相等. 理由:根据题意得;()2

,1n C k n k

nk ----+,()211,1n C k n k nk k ---+--++,

∴C n C n –1两点之间的铅直高度=()22

11k nk k k nk k --++---+=.

C n C n –1两点之间的水平距离=1()1k n k n --+---=. ∴由勾股定理得C n C n –12=k 2+1, ∴C n C n –121k +.

③n

n

C A 与1

1n n C

A --不平行.

理由:

根据题意得:()2

,1n C k n k

nk ----+,()211,1n C k n k nk k ---+--++,

()

,1n A n -,()1

1,1n A n --+.

过C n ,C n –1分别作直线y =1的垂线,垂足为D ,E ,

所以D (–k –n ,1),E (–k –n +1,1). 在Rt △DA n C n 中,

tan ∠DA n C n =()2

211()n n k nk C D k nk

k n A D n k n k

---++===+----, 在Rt △EA n –1C n –1中,

tan ∠EA n –1C n –1=

()2

2111111(1)n n k nk k C E k nk k

k n A E n k n k

-----+++-===+--+---+, ∵1k n +-≠k n +,

∴tan ∠DA n C n ≠tan ∠EA n –1C n –1, ∴n

n

C A 与1

1n n C

A --不平行.

9.(2019·甘肃白银)阅读下面的例题及点拨,并解决问题: 例题:如图①,在等边△ABC 中,M 是BC 边上一点(不含端点B ,C ),N 是△ABC 的外角∠ACH 的平分线上一点,且AM =MN .求证:∠AMN =60°.

点拨:如图②,作∠CBE =60°,BE 与NC 的延长线相交于点E ,得等边△BEC ,连接EM .易证:△ABM ≌△EBM (SAS ),可得AM =EM ,∠1=∠2;又AM =MN ,则EM =MN ,可得∠3=∠4;由∠3+∠1=∠

4+

∠5=60°,进一步可得∠1=∠2=∠5,又因为∠2+∠6=120°,所以∠5+∠6=120°,即:∠AMN=60°.

问题:如图③,在正方形A1B1C1D1中,M1是B1C1边上一点(不含端点B1,C1),N1是正方形A1B1C1D1的外角∠D1C1H1的平分线上一点,且A1M1=M1N1.求证:∠A1M1N1=90°.

【答案】见解析.

【解析】延长A1B1至E,使EB1=A1B1,连接EM1、EC1,

如图所示:

则EB1=B1C1,∠EB1M1=90°=∠A1B1M1,

∴△EB1C1是等腰直角三角形,

∴∠B1EC1=∠B1C1E=45°,

∵N1是正方形A1B1C1D1的外角∠D1C1H1的平分线上一点,∴∠M1C1N1=90°+45°=135°,

∴∠B1C1E+∠M1C1N1=180°,

∴E、C1、N1三点共线,

在△A1B1M1和△EB1M1中,

111

11111

1111

A B EB

A B M EB

M

M

B M B

=

?

?

∠=∠

?

?=

?

∴△A1B1M1≌△EB1M1(SAS),

∴A1M1=EM1,∠1=∠2,

∵A1M1=M1N1,∴EM1=M1N1,∴∠3=∠4,

∵∠2+∠3=45°,∠4+∠5=45°,∴∠1=∠2=∠5,

∵∠1+∠6=90°,∴∠5+∠6=90°,

∴∠A1M1N1=180°﹣90°=90°.

【名师点睛】此题是四边形综合题目,考查了正方形的性质、全等三角形的判定与性质、等腰直角三角形的判定与性质、等腰三角形的判定与性质、三角形的外角性质等知识;本题综合性强,熟练掌握正方形的性质,通过作辅助线构造三角形全等是解本题的关键.10.(2019·甘肃天水)如图1,对角线互相垂直的四边形叫做垂美四边形.

(1)概念理解:如图2,在四边形ABCD中,AB=AD,CB=CD,问四边形ABCD是垂美四边形吗?请说明理由;

(2)性质探究:如图1,四边形ABCD的对角线AC、BD交于点O,AC⊥BD.

试证明:AB2+CD2=AD2+BC2;

(3)解决问题:如图3,分别以Rt△ACB的直角边AC和斜边AB 为边向外作正方形ACFG和正方形ABDE,连结CE、BG、GE.已知AC=4,AB=5,求GE的长.

【答案】(1)四边形ABCD是垂美四边形.理由见解析.(2)见解析.(3)GE73

【解析】(1)四边形ABCD是垂美四边形.理由如下:

∵AB=AD,∴点A在线段BD的垂直平分线上,

∵CB=CD,∴点C在线段BD的垂直平分线上,

∴直线AC是线段BD的垂直平分线,

∴AC⊥BD,即四边形ABCD是垂美四边形;

(2)如图1,

∵AC⊥BD,∴∠AOD=∠AOB=∠BOC=∠COD=90°,

由勾股定理得,AB2+CD2=AO2+BO2+DO2+CO2=AD2+BC2,

∴AD2+BC2=AB2+CD2;

(3)连接CG、BE,

∵∠CAG=∠BAE=90°,

∴∠CAG+∠BAC=∠BAE+∠BAC,即∠GAB=∠CAE,

在△GAB和△CAE中,

AG AC

GAB CAE AB AE

=

?

?

∠=∠

?

?=

?

∴△GAB≌△CAE(SAS),

∴∠ABG=∠AEC,又∠AEC+∠AME=90°,

∴∠ABG+∠AME=90°,即CE⊥BG,

∴四边形CGEB是垂美四边形,

由(2)得,CG2+BE2=CB2+GE2,

∵AC=4,AB=5,∴BC=3,CG2,BE2,

∴GE2=CG2+BE2-CB2=73,∴GE73

【名师点睛】(1)根据垂直平分线的判定定理证明即可;(2)根据垂直的定义和勾股定理解答即可;

(3)根据垂美四边形的性质、勾股定理、结合(2)的结论计算.本题考查的是正方形的性质、全等三角形的判定和性质、垂直的定义、勾股定理的应用,正确理解垂美四边形的定义、灵活运用勾股定理是解题的关键.

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