Unit 4 I used to be afraid of the dark Section B 3a-Self Check 课时练 2025-2026学年上学期初中英语人教版九年级全册

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名称 Unit 4 I used to be afraid of the dark Section B 3a-Self Check 课时练 2025-2026学年上学期初中英语人教版九年级全册
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中小学教育资源及组卷应用平台
Unit 4 I used to be afraid of the dark Section B 3a-Self Check 课时练
2025-2026学年上学期初中英语人教版九年级全册
一、翻译
将下列短语翻译成英语。
1.过去是/常做
2.习惯于做
3.害怕
4.喜欢做……
5.间或;偶尔
6.处理
7.为……感到骄傲
二、单词语法填空
8.The girl doesn’t dare (speak) in front of the class.
9.There are two days left. Haven’t you made a (decide) yet
10.Our Chinese teacher is very (humor). His classes are very funny.
11.Everyone is required (show) their tickets before entering the hall.
12.I hope this excellent report will be (help) in answering your question.
三、选词填空
silent be proud of in person influence humorous
13.The mother traveled for many hours to return home to talk to her child .
14.Kate’s grandparents have had a great on her.
15.That British teacher is very . He always tells us interesting jokes.
16.He used to be a very quiet teenager. He remained most of the time.
17.Tina played very well in the basketball game and her parents her.
四、 将下列句子翻译成英语。
18.她不再害怕在众人面前唱个歌了。(not…anymore)
19.他过去讨厌体育课。(used to do)
20.最近几年发生了很多变化。(take place)
21.她的祖父母对她有很大的影响。(make)
五、阅读理解
Do you know when restaurants were invented The first restaurants were invented thousands of years ago. However, they were very different from restaurants today.
The first restaurants were in ancient Greece(希腊) and ancient Rome. These restaurants served food in large stone bowls. People didn’t order food from a menu. Everyone shared the food from big bowls. People believe that these places were very popular because most homes in ancient Greece and Rome did not have kitchens. Also, people didn’t have to store(储存) food at home if they ate their meals at these restaurants.
Later, restaurant began to open in China. In the early 1100s, more than one million people lived in the city of Hangzhou, China. It was a very busy city, and people had money. All of these people had to eat. Smart cooks started cooking and selling food along the big street in the city. Unlike in ancient Greece and Rome, people in China could choose food from a menu. They didn’t have to eat the same food as everyone else.
For the next several centuries, there were restaurants all over the world. People could buy food in the street or at inns(小旅馆). Then in the middle of the 1700s, restaurants started opening in Paris. These restaurants were more similar to restaurants we know today. There was a wider choice of food, and eating in these restaurants was a more enjoyable experience. In the 19th century, trains made travel much faster and simpler. In the end, this kind of restaurant began to appear all over Europe and in other parts of the world.
Now, restaurants are everywhere. You can buy different kinds of food. You can get food sent to your door. But remember it wasn’t always like that.
22.It is believed that these first restaurants were popular because ________.
A.people could eat different food B.the food there was very cheap
C.people were too busy to cook D.most homes didn’t have kitchens
23.How were the restaurants in Hangzhou different from the first restaurants
A.They offered menus to customers.
B.They usually served the same food.
C.They sold the food in large stone bowls.
D.They prepared food for all people in the city.
24.What does the underlined part “these restaurants” in Paragraph 4 refer to
A.Restaurants in ancient Greece and ancient Rome.
B.Restaurants in Hangzhou in the early 1100s.
C.Restaurants in Paris in the middle of the 1700s.
D.Restaurants all over Europe in the 19th century.
25.Which of the following is TRUE according to the passage
A.Now people can get food without going out.
B.The first restaurants were similar to restaurants today.
C.Hangzhou used to be a city that had no smart cooks.
D.People in ancient Greece and Rome usually stored food at home.
六、书面表达
26.杂志社正在开展主题为“今昔变化”的征文活动,请你结合以下要点用英语写一篇题为“Great changes in communication”的短文参加此次活动。
要点:1. 过去写信或使用公共电话,现在使用手机或因特网;
2. 过去接收信息要花费很长时间,现在花费几秒钟;
3. 这些变化改变着人们的生活。
注意:1.短文须包括提示中的两个要点和一个补充要点,可适当发挥。
2.短文中不要出现真实的地名、校名和人名。
3.字数80—100词左右。(开头已给出,不计入总词数)
Great changes in communication
Now the development of science and technology is becoming faster and faster, so the way people communicate with each other is becoming more and more convenient.
__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
参考答案
题号 22 23 24 25
答案 D A C A
1.used to be/do 2.be/get used to doing 3.be afraid of 4.like to do/like doing 5.from time to time 6.deal with 7.be proud of/take pride in
【解析】1.过去是/常做:used to be/do,动词短语。故填used to be/do。
2.习惯于做:be/get used to doing,动词短语。故填be/get used to doing。
3.害怕:be afraid of,形容词短语。故填be afraid of。
4.喜欢做……:like to do/like doing,动词短语。故填like to do/like doing。
5.间或;偶尔:from time to time,介词短语。故填from time to time。
6.处理:deal with,动词短语。故填deal with。
7.为……感到骄傲:be proud of,形容词短语;take pride in,动词短语。故填be proud of/take pride in。
8.to speak
【详解】句意:这个女孩不敢在全班同学面前讲话。speak“说话”,是一个动词。dare to do sth.“敢做某事”,固定搭配。故填to speak。
9.decision
【详解】句意:还剩两天。你还没有做决定吗?decide“决定”,动词;根据“made a...”可知,此处考查make a decision“做决定”。故填decision。
10.humorous
【详解】句意:我们的语文老师很幽默。他的课很有趣。根据“His classes are very funny.”可知,此处是指老师很幽默;应用名词humor的形容词形式humorous“幽默的”作表语。故填humorous。
11.to show
【详解】句意:每个人在进入大厅前都必须出示门票。根据空前“is required”可知,此处考查be required to do sth.“被要求做某事”,固定短语,应用动词不定式作主语补足语。故填to show。
12.helpful
【详解】句意:我希望这份出色的报告在回答你的问题上会有帮助。此处在句中作表语,用形容词helpful“有帮助的”,故填helpful。
13.in person 14.influence 15.humorous 16.silent 17.were proud of
【解析】13.句意:这位母亲旅行了很多小时回家亲自和她的孩子谈话。根据“The mother traveled for many hours to return home to talk to her child”以及结合选词,可知是指亲自和她的孩子谈话。in person意为“亲自”,符合句意。故填in person。
14.句意:凯特的祖父母对她有很大的影响。have a great influence on sb.意为“对某人有很大的影响”,故填influence。
15.句意:那位英国老师非常幽默。他总是给我们讲有趣的笑话。根据“He always tells us interesting jokes.”可知是指那位英国老师非常幽默。humorous意为“幽默的”,故填humorous。
16.句意:他过去是一个非常安静的青少年。他大部分时间都保持沉默。根据“He used to be a very quiet teenager.”可知是指他大部分时间都保持沉默。silent意为“沉默的”,remained是系动词,后接形容词修饰。故填silent。
17.句意:蒂娜在篮球比赛中打得非常好,她的父母为她感到骄傲。根据“Tina played very well in the basketball game”可知是指她的父母为她感到骄傲。be proud of意为“为……感到骄傲”,时态为一般过去时,主语是复数parents,be动词用过去式were。故填were proud of。
18.She wasn’t afraid to sing in public anymore. 19.He used to hate P.E. class. 20.Great changes have taken place in the past few years. 21.Her grandparents have made a great difference to her.
【解析】18.分析题干可知,句子为陈述句,且时态为一般过去时;she“她”,作主语,位于句首,首字母要大写;be afraid to do sth.“害怕做某事”,形容词短语,主语为“she”,be动词用was;not…anymore“不再”,副词短语;sing“唱歌”,动词;in public“在公共场合”,介词短语。故填She wasn’t afraid to sing in public anymore.
19.分析题干可知,句子为陈述句,且时态为一般过去时;he“他”,作主语,位于句首,首字母要大写;used to do sth.“过去常常做某事”,动词短语;hate P.E. class“讨厌体育课”。故填He used to hate P.E. class.
20.分析句子结构可知,句子为陈述句,且时态为现在完成时;great changes“巨大的变化”,作主语,句首首字母要大写;take place“发生”,动词短语;现在完成时的结构为have/has done,主语为第三人称复数,助动词用have,take的过去分词为taken;in the past few years“在过去的几年中”,介词短语。故填Great changes have taken place in the past few years.
21.分析句子结构可知,句子为陈述句,且时态为现在完成时;her grandparents“她的祖父母”,作主语,句首首字母要大写;make a great difference to sb.“对某人有很大影响”,动词短语;现在完成时的结构为have/has done,主语为第三人称复数,助动词用have,动词make的过去分词为made;her“她”,人称代词宾格,作介词to的宾语。故填Her grandparents have made a great difference to her.
22.D 23.A 24.C 25.A
【导语】本文主要介绍了餐馆的发展史。
22.细节理解题。根据第二段第四句“People believe that these places were very popular because most homes in ancient Greece and Rome did not have kitchens.”可知,人们认为这些餐馆很受欢迎,因为古希腊和罗马的大多数家庭都没有厨房。故选D。
23.细节理解题。根据第三段倒数第二句“Unlike in ancient Greece and Rome, people in China could choose food from a menu.”可知,与古希腊和古罗马不同,中国人可以从菜单中选择食物。故选A。
24.词句猜测题。根据第四段“Then in the middle of the 1700s, restaurants started opening in Paris. These restaurants were more similar to restaurants we know today.”可知,These restaurants指18世纪中叶的巴黎餐馆。故选C。
25.推理判断题。根据最后一段“You can get food sent to your door.”可知你可以让食物送到你的家门口,故选项A正确;根据第一段“The first restaurants were invented thousands of years ago. However, they were very different from restaurants today.”以及第二段内容可知,第一家餐馆与今天的很不一样,故选项B错误;根据第三段“Smart cooks started cooking and selling food along the big street in the city.”可知杭州过去就有聪明的厨师,故选项C错误;根据第二段“Also, people didn’t have to store food at home if they ate their meals at these restaurants.”可知,古希腊和罗马人没必要储存食物,故选项D错误。故选A。
26.例文:
Great changes in communication
Now the development of science and technology is becoming faster and faster, so the way people communicate with each other is becoming more and more convenient.
In the past, people used to keep in touch with each other mainly by writing letters or using the public telephone. But now people can communicate with each other anywhere at anytime on their mobile phones. Besides they can send and receive messages on QQ or WeChat. It used to take several days or even weeks to hear from each other, but now it takes only several seconds, even if they’re in different countries.
With these changes, people’s life has improved a lot. They not only know more about the outside world, but they also have a more comfortable life.
【详解】1.题干解读:该题目属于议论文写作。在写作时应紧接开头继续写作,根据题目要求讲述在人际交往的方式上过去和现在的变化以及这些变化对人们的生活的影响。
2.写作指导:本文时态以一般现在时和一般过去时为主,人称以第三人称为主,讲述过去和现在人际交往上的变化以及对人们的生活的影响,要叙述完整,条理清晰。
3.2 函数的基本性质--函数的单调性和最大(小)值 常见题型总结练 2025-2026学年数学高一年级人教A版(2019)必修第一册
一:图象法求单调区间
1.如图是函数的图象,则函数的单调递减区间为( )
A. B. C. D.
2.函数的单调递增区间是( )
A. B. C. D.
3.已知函数的图象如图所示,则该函数的减区间为( )

A. B.
C. D.
4.定义在上的函数的单调递减区间是 .
二:函数单调性的判断
1.已知四个函数的图象如图所示,其中在定义域内具有单调性的函数是( )
A. B.
C. D.
2.(多选题)在区间上为减函数的是( )
A. B. C. D.
3.(多选题)下列函数中,在R上是增函数的是( )
A.y=|x| B.y=x
C.y=x2 D.y=
4.下列函数中,在上单调递增的是( )
A. B. C. D.
三:证明或判断函数的单调性
1.下列函数中,满足“对任意,,当时,都有”的是( )
A. B. C. D.
2.函数在上的最小值为( )
A.1 B. C. D.
3.下列函数中,在区间上为增函数的是( )
A. B. C. D.
4.已知函数的定义域为,则下列说法中正确的是( )
A.若满足,则在区间内单调递增
B.若满足,则在区间内单调递减
C.若在区间内单调递增,在区间内单调递增,则在区间内单调递增
D.若在区间内单调递增,在区间内单调递增,则在区间内单调递增
四:求函数的单调区间
1.函数的单调增区间为( )
A. B. C.和 D.
2.函数的单调递增区间是( )
A.(,1] B.[1,) C.[1,4] D.[2,1]
3.已知,则函数的单调增区间是 .
4.(24-25高一上·全国·课堂例题)已知函数,,根据图象写出它的单调区间..
五:函数单调性的应用
1.已知函数在区间上是减函数,则整数a的取值可以为( )
A. B. C.0 D.1
2.若函数在区间上单调递减,则实数的取值范围是( )
A. B. C. D.
3.若函数(为实数)是R上的减函数,则( )
A. B. C. D.
4.若在上为减函数,则实数的取值范围为( )
A. B. C. D.
六:利用单调性比较大小或解不等式
1.若函数在上单调递增,且,则实数的取值范围是( )
A. B. C. D.
2.已知函数f(x)的定义域为R,且对任意的x1,x2且x1≠x2都有[f(x1)﹣f(x2)](x1﹣x2)>0成立,若f(x2+1)>f(m2﹣m﹣1)对x∈R恒成立,则实数m的取值范围是(  )
A.(﹣1,2) B.[﹣1,2]
C.(﹣∞,﹣1)∪(2,+∞) D.(﹣∞,﹣1]∪[2,+∞)
3.设函数在区间上有意义,任意两个不相等的实数,下列各式中,能够确定函数在区间上单调递增的是( )
A. B.
C. D.
4.(多选题)设函数在上为减函数,则( )
A.
B.
C.
D.
E.
函数的最大(小)值
一:利用图象求函数最值
1.定义在R上的偶函数在[0,7]上是增函数,在[7,+∞)上是减函数,又f(7)=6,则f(x)(  )
A.在[-7,0]上是增函数,且最大值是6
B.在[-7,0]上是减函数,且最大值是6
C.在[-7,0]上是增函数,且最小值是6
D.在[-7,0]上是减函数,且最小值是6
2.函数y=f(x)在[-2,2]上的图象如图所示,则此函数的最小值、最大值分别是(  ).
A.f(-2),0 B.0,2 C.f(-2),2 D.f(2),2
3.若函数,它的最大值为,则实数的取值范围是( )
A. B. C. D.
4.函数在区间上的值域为
二:利用单调性求函数最值
1.函数y=在[2,3]上的最小值为( )
A.2 B.
C. D.-
2.已知函数在区间上的最大值为A,最小值为B,则A-B等于( )
A. B. C.1 D.-1
3.函数在区间上的最小值为( )
A. B.1 C. D.2
4.若函数y=在区间[2,4]上的最小值为5,则k的值为(  )
A.5 B.8
C.20 D.无法确定
三:求二次函数的最值
1.已知函数在区间上有最大值5,最小值1,则的值等于( )
A. B.1 C.2 D.3
2.定义域为R的函数满足,且当时,,则当时,的最小值为(  )
A. B. C. D.
3.(多选题)关于函数()在上最小值的说法不正确的是( )
A.4 B.
C.与的取值有关 D.不存在
4.(多选题)已知在区间上的最小值为,则可能的取值为( )
A. B.3 C. D.1
四:判断二次函数的单调性和求解单调区间
1.函数在区间上递增,则实数的取值范围是(  )
A. B. C. D.
2.若函数在上是减函数,则实数m的取值范围是( )
A. B. C. D.
3.若函数在上是减函数,则实数m的取值范围是( )
A. B. C. D.
4.(多选题)已知函数的定义域为,值域为,则的可能的取值是( )
A.1 B.2 C.3 D.4
五:函数最值的实际应用
1.如图所示是函数的图象,图中曲线与直线无限接近但是永不相交,则以下描述正确的是( )
A.函数的定义域为
B.函数的值域为
C.此函数在定义域中不单调
D.对于任意的,都有唯一的自变量x与之对应
2.若是偶函数,且对任意∈且,都有,则下列关系式中成立的是( )
A. B.
C. D.
3.向一个圆台形的容器(如图所示)中倒水,且任意相等的时间间隔内所倒的水体积相等,记容器内水面的高度y随时间t变化的函数为,则以下函数图象中,可能是的图象的是(  ).
A. B.
C. D.
4.(23-24高一上·全国·课后作业)一水池有2个进水口,1个出水口,进出水速度如图甲、乙所示.某天0点到6点,该水池的蓄水量如图丙所示(至少打开一个水口).

给出以下4个论断,其中正确的是(  )
A.0点到3点只进水不出水
B.3点到4点不进水只出水
C.3点到4点只有一个进水口进水
D.4点到6点不进水也不出水
答案
一:图象法求单调区间
根据题意,结合函数图象可得函数的单调递减区间为:.
故选:.
函数的定义域需要满足,解得定义域为,
因为在上单调递增,所以在上单调递增,
故选:D.
函数的图象在区间和是下降的,在区间和是上升的,
故该函数的减区间为.
故选:C.
,取
如图所示:
单调递减区间是
故答案为
二:函数单调性的判断
对于A,函数分别在及上单调递增,
但存在,使,故A不符合题意;
对于C,函数分别在及上单调递增,
但存在,使,故C不符合题意;
对于D,函数分别在及上单调递减,
但存在,,使,故D不符合题意;
只有B完全符合增函数的定义,具有单调性.
故选:B.
解:函数是上的减函数,
函数在区间上单调递减,
函数在区间单调递减.
函数在区间单调递增,
所以A,B,C符合要求;D项不符合要求.
故选:ABC.
解:选项A,,当x<0时单调递减,不符合题意;
选项B,显然在R上是增函数,符合题意;
选项C,y=x2,当x<0时单调递减,不符合题意;
选项D,作出草图如下,实线部分,观察图象可得函数在R上为增函数,符合题意.

故选:BD
对于A中,函数在上单调递减,所以A不符合题意;
对于B中,函数在上单调递减,单调递增,所以B符合题意;
对于C中,函数在上单调递减,所以C不符合题意;
对于D中,时函数在上单调递减,所以D符合题意.
故选:D.
三:证明或判断函数的单调性
因为对任意,,当时,都有,所以在上为增函数,
A选项,在上为增函数,不符合题意.
B选项,在上为减函数,不符合题意.
C选项,在上为增函数,符合题意.
D选项,在上为增函数,不符合题意.
故选:C.
因为在上单调递增,且恒成立,
可知函数在上单调递减,
当时,,所以函数在上的最小值为.
故选:B.
选项A:,开口向下,对称轴为,所以函数在区间上为减函数,故选项A错误;
选项B:,所以函数在区间上为增函数,故选项B正确;
选项C:可以看作由函数向左平移一个单位得到,所以函数在区间上为减函数,故选项C错误;
选项D:,开口向下,对称轴为,所以函数在区间上为减函数,故选项D错误.
故选:B.
对于AB:函数满足,或,特值并不具有任意性,
所以区间端点值的大小关系并不能确定函数在区间上的单调性,故A,B错误;
对于C:区间和有交集,故函数在区间内单调递增,故C正确,
对于D:区间和没有交集,故不能确定函数在区间内的单调性.
例如在和上递增,但,故D错误.
故选:C.
四:求函数的单调区间
由可得且,
因为开口向下,其对称轴为,
所以的减区间为和
所以的单调增区间为和
故选:C
由,得,解得,
令,则,
因为在上递增,在上递减,而在上递增,
所以在上递增,在上递减,
所以的单调递增区间是,
故选:D
解:因为,对称轴为 ,又开口向下,
又,∴函数的单调递增区间为.
故答案为:

函数图象如图所示.
由图象可知,函数的单调递增区间为,单调递减区间为.
五:函数单调性的应用
解:由题意可得,解得,
∴整数a的取值可以为.
故选:A
函数的对称轴为,
由题意可知,解得,
所以实数的取值范围是.
故选:B.
由题意知,解得
故选:D
为上的减函数, 时, 递减,即,①, 时, 递减,即,②且 ,③ 联立①②③解得, .
故选:C.
六:利用单调性比较大小或解不等式
在上单调递增,,,解得:,
实数的取值范围为.
故选:C.
解:由题意,可知:
∵对任意的x1,x2且x1≠x2都有[f(x1)﹣f(x2)](x1﹣x2)>0成立,
∴函数f(x)在定义域R上为增函数.
又∵f(x2+1)>f(m2﹣m﹣1)对x∈R恒成立,
∴x2+1>m2﹣m﹣1,
∴m2﹣m﹣1<1,
即:m2﹣m﹣2<0.
解得﹣1<m<2.
故选:A.
解:函数在区间上单调递增,则任意两个不相等的实数,与应该同号,所以,
故选:C.
由题意,函数在上为减函数.
当时,,,,
则,,,故ACD错误;
对于B,因为,所以,
所以,故B正确;
对于E,因为,所以,故E正确.
故选:BE.
函数的最大(小)值
一:利用图象求函数最值
∵函数是偶函数,而且在[0,7]上为增函数,
∴函数在[-7,0]上是减函数.
又∵函数在x=7和x=-7的左边是增函数,右边是减函数,且f(7)=f(-7),
∴最大值为f(7)=f(-7)=6.
故选B.
试题分析:由图观察可知函数在和上单调递增,在上单调递减.
所以函数在处取的最大值为.
又由图观察可知,所以函数的最小值为.故C正确.
由题意,函数表示开口向上,且对称轴为的抛物线,
要使得当,函数的最大值为,则满足且,
解得,所以实数的取值范围是.
故选D.
由题:,函数在单调递减,在单调递减,
可以看成函数向右平移1个单位,再向上平移1个单位,作出图象:
所以函数在递减,在递减,,,
所以函数的值域为.
故答案为:
二:利用单调性求函数最值
y=在[2,3]上单调递减,所以x=3时取最小值为,
故选:B.
函数在区间是减函数,
所以时有最大值为1,即A=1,
时有最小值,即B=,
则,
故选:A.
由知,在上是增函数,所以在上递增,所以.
故选:C
∴或∴k=20.选C.
三:求二次函数的最值
由题意,函数,
可得函数在区间上单调递增,在区间上单调递减,
当时,则函数在区间上单调递增,其最小值为,
显然不合题意;
当时,则函数在区间上单调递增,在区间上单调递减,
故函数的最大值为,
因为,令,即,即,
解得或,
又因为,所以.
故选: D.
设,则,则,又,∴,∴当时,取到最小值为.
由题意得:二次函数()的对称轴为,且函数图象开口向上,
则该函数在上单调递减,
所以,
故选:BCD.
解:因为函数,函数的对称轴为,开口向上,
又在区间上的最小值为,
所以当时,,解得(舍去)或;
当,即时,,解得(舍去)或;
当,即时,.
综上,的取值集合为.
故选:BC.
四:判断二次函数的单调性和求解单调区间
函数,二次函数图像开口向上,
若在区间上递增,
则对称轴x=-a,
即a
故选D.
函数的对称轴为,
由于在上是减函数,所以.
故选:B
函数的对称轴为,
由于在上是减函数,所以.
故选:B
因为函数在区间上单调递减,在上单调递增,
所以在R上的最小值为,且,
(1)当时,由的值域为,可知必有
所以且,解得,此时
(2)当时,由的值域为,可知必有
所以且,解得,此时
综上可知,
所以的可能的取值为
故选:BCD
五:函数最值的实际应用
1 由图知:的定义域为,值域为,A、B错;
显然在分别递增,但在定义域上不单调,C对;
显然,对应自变量x不唯一,D错.
故选:C
∵对任意的x1,x2∈(0,+∞),都有,
∴函数f(x)在(0,+∞)上单调递减,
又∵,
∴,
又∵f(x)是偶函数,∴f(﹣)=f().
∴.
故选:A.
由容器的形状可知,在相同的变化时间内,高度的增加量越来越小,
故函数的图象越来越平缓,
故选:D.
由甲,乙图得进水速度为1,出水速度为2,
对A,由题意可知在0点到3点这段时间,每小时进水量为2,即2个进水口同时进水且不出水,所以A正确;
对BC,从丙图可知3点到4点水量减少了1,所以应该是有一个进水口进水,同时出水口也出水,故B错误C正确;
对D,当两个进水口同时进水,出水口也同时出水时,水量保持不变;也可由题干中的“至少打开一个水口”知D错,故D错误.
故选:AC
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