当前位置:文档之家› 9 第七章 第8节 随堂演练巩固提升

9 第七章 第8节 随堂演练巩固提升

9 第七章 第8节 随堂演练巩固提升
9 第七章 第8节 随堂演练巩固提升

[随堂检测]

1.在下列几种运动中,遵守机械能守恒定律的有( )

A .雨点匀速下落

B .平抛运动

C .汽车刹车时的运动

D .物体沿斜面匀速下滑

解析:选B.机械能守恒的条件是只有重力做功.A 中除重力外,有阻力做功,机械能不守恒;B 中只有重力做功,机械能守恒;C 中有阻力做功,机械能不守恒;D 中物体除受重力外,有阻力做功,机械能不守恒.

2.如图,半圆形光滑轨道固定在水平地面上,半圆的直径与地面垂直.一小物块以速度v 从轨道下端滑入轨道,并从轨道上端水平飞出,小物块落地点到轨道下端的距离与轨道半径有关,此距离最大时对应的轨道半径为(重力加速度大小为g )( )

A .v 216g

B .v 28g

C .v 2

4g D .v 22g

解析:选B.设轨道半径为R ,小物块从轨道上端飞出时的速度为v 1,由于轨道光滑,根

据机械能守恒定律有mg ×2R =12m v 2-12m v 21

,小物块从轨道上端飞出后做平抛运动,对运动分解有:x =v 1t ,2R =12gt 2,求得x =-16????R -v 28g 2+v 44g 2,因此当R -v 28g =0,即R =v 2

8g 时,x 取得最大值,B 项正确,A 、C 、D 项错误.

3.取水平地面为重力势能零点.一物块从某一高度水平抛出,在抛出点其动能与重力势能恰好相等.不计空气阻力.该物块落地时的速度方向与水平方向的夹角为( )

A .π6

B .π4

C .π3

D .5π12

解析:选B.设物块水平抛出的初速度为v 0,高度为h ,由题意知12m v 20

=mgh ,即v 0=2gh .物块在竖直方向上的运动是自由落体运动,故落地时的竖直分速度v y =2gh =v x =v 0,则该

物块落地时的速度方向与水平方向的夹角θ=π4

,故选项B 正确,选项A 、C 、D 错误. 4.如图所示,长为L 的轻绳一端固定于O 点,另一端系一质量为m 的小球.现将绳水平拉直,让小球从静止开始运动,重力加速度为g ,当绳与竖直方向的夹角α=30°时,小球受到的合力大小为( )

A .3mg

B .132mg

C .32mg

D .(1+3)mg

解析:选B.由机械能守恒定律可知,在与竖直方向夹角为30°时,mgL cos α=12

m v 2,结合圆周运动向心力公式F 向=m v 2L =3mg ,沿轻绳方向,F 向=F T -mg cos α,解得F T =332

mg ,由正交分解法把轻绳拉力及重力在水平、竖直方向分解,水平方向的合力为F T sin α=334

mg ,竖直方向的合力为F T cos α-mg =54

mg ,由勾股定理可知,选项B 正确. 5.如图所示,在光滑的水平桌面上放置一根长为l 的链条,链条

沿桌边挂在桌外的长度为a ,链条由静止开始释放,求链条全部离开桌

面时的速度是多少?

解析:当链条从图示位置到全部离开桌面的过程中,原来桌面上的

那段链条重心下降的距离为l -a 2

,挂在桌边的那段链条重心下降的距离为l -a ,设链条单位长度的质量为m ′,链条总的质量为m =lm ′,则ΔE k =ΔE p ,即m ′(l -a )g

l -a 2+m ′ag (l -a )=12lm ′v 2,解得v = (l 2-a 2)g l . 答案: (l 2-a 2)g l

6.如图所示,质量为m =2 kg 的小球系在轻弹簧的一端,另一端固

定在悬点O 处,将弹簧拉至水平面位置A 处,且弹簧处于自然状态,由

静止释放,小球到达距O 点下方h =0.5 m 处的B 点时速度为v =2 m/s .求

小球从A 运动到B 的过程中弹簧的弹力做的功.(g 取10 m/s 2)

解析:小球在由A 至B 的过程中,只受重力和弹力作用,故系统的机械能守恒.以B 点为参考平面,则在初状态A ,系统的动能E k1=0

重力势能E p1=mgh

机械能E 1=E k1+E p1=mgh

在末状态B ,系统的动能E k2=m v 22

设(弹性)势能为E p2,机械能为E 2=E k2+E p2=m v 22

+E p2 对系统在运动过程的初、末状态,由机械能守恒定律有mgh =m v 22

+E p2 所以E p2=mgh -m v 22=2×10×0.5 J -2×222

J =6 J 因为弹性势能增加,弹簧的弹力做负功,故弹簧的弹力做的功为-6 J .

答案:-6 J

[课时作业]

一、单项选择题

1.下列运动过程满足机械能守恒的是( )

A .电梯匀速下降过程

B .起重机吊起重物过程

C .物体做自由落体运动过程

D .考虑阻力条件下滑雪者沿斜面下滑过程

解析:选C.机械能守恒的条件是只有重力做功,只发生动能和势能的转化,电梯匀速下降过程中,重力做正功,拉力做负功,机械能减少,A 错误;起重机吊起重物过程,重力做负功,拉力做正功,机械能增加,B 错误;物体做自由落体运动过程,只有重力做功,机械能守恒,C 正确;考虑阻力条件下滑雪者沿斜面下滑过程,重力做正功,支持力不做功,滑动摩擦力做负功,机械能减少,D 错误.

2.如图所示,将一个内、外侧均光滑的半圆形槽置于光滑的水平面

上,槽的左侧有一竖直墙壁.现让一小球自左端槽口A 点的正上方由静

止开始下落,从A 点与半圆形槽相切进入槽内,则下列说法正确的是( )

A .小球在半圆形槽内运动的全过程中,只有重力对它做功

B .小球从A 点向半圆形槽的最低点运动的过程中,小球处于失重状态

C.小球从A点经最低点向右侧最高点运动的过程中,小球与槽组成的系统机械能守恒D.小球从下落到从右侧离开槽的过程机械能守恒

解析:选C.小球从A点向半圆形槽的最低点运动的过程中,半圆形槽有向左运动的趋势,但是实际上没有动,整个系统只有重力做功,所以小球与槽组成的系统机械能守恒.而小球过了半圆形槽的最低点以后,半圆形槽向右运动,由于系统没有其他形式的能量产生,满足机械能守恒的条件,所以系统的机械能守恒.小球从开始下落至到达槽最低点前,小球先失重,后超重.当小球向右上方滑动时,半圆形槽也向右移动,半圆形槽对小球做负功,小球的机械能不守恒.综合以上分析可知选项C正确.

3.如图所示,A、B两球的质量相同,A球系在不可伸长的绳上,B

球固定在轻质弹簧上,把两球都拉到水平位置(绳和弹簧均拉直且为原长),

然后释放.当小球通过悬点O正下方的C点时,弹簧和绳子等长,则此时

( )

A.A、B两球的动能相等

B.A球重力势能的减少量大于B球重力势能的减少量

C.A球所在系统的机械能大于B球所在系统的机械能

D.A球的速度大于B球的速度

解析:选D.A球运动过程中,仅有重力对其做功,B球运动过程中,仅有重力和弹簧弹力对其做功,故A、B球所在系统的机械能均守恒.以过C的水平面为零势能面,A、B球在运动过程中重力做功相同,重力势能的减少量相同,但B球有一部分重力势能转化为弹簧的弹性势能,所以到达C点时A球的动能大,速度大,只有D正确.

4.将物体从地面竖直上抛,如果不计空气阻力,物体能够达到的最大高度为H.当物体在上升过程中的某一位置,它的动能是重力势能的3倍,则这一位置的高度是( ) A.2H/3 B.H/2

C.H/3 D.H/4

解析:选D.物体在运动过程中机械能守恒,设动能是重力势能的3倍时的高度为h,取地面为零势能面,则有mgH=E k+mgh,即mgH=4mgh,解得:h=H/4,故D正确.5.如图所示,竖直立在水平地面上的轻弹簧,下端固定在地面上,将

一个金属球放置在弹簧顶端(球与弹簧不拴接),并用力向下压球,使弹簧压

缩(在弹性限度内)一定程度后,用竖直细线把弹簧拴牢.现突然烧断细线,

球将被弹起,脱离弹簧后能继续向上运动,那么从细线被烧断到金属球刚脱离弹簧的过程中,

下列说法正确的是( )

A.金属球的机械能守恒

B.金属球的动能一直在减少,而机械能一直在增加

C.在刚脱离弹簧的瞬间金属球的动能最大

D.金属球的动能与弹簧的弹性势能之和一直在减少

解析:选D.烧断细线后,开始的一段时间内,弹簧弹力大于金属球的重力,金属球向上做加速运动,当弹簧的弹力小于金属球的重力后,金属球向上做减速运动,因此当重力与弹力相等时,金属球的速度最大,在整个运动过程中,金属球、弹簧组成的系统机械能守恒,金属球向上运动的过程中,弹簧的弹性势能减少,金属球的机械能一直增加,故选项A、B、C错误;金属球与弹簧组成的系统机械能守恒,金属球的动能、重力势能、弹簧的弹性势能之和保持不变,金属球向上运动的过程中,金属球的重力势能一直增加,所以金属球的动能和弹簧的弹性势能之和一直减少,选项D正确.

6.如图所示,有一内壁光滑的闭合椭圆形管道,置于竖直平面内,

MN是通过椭圆中心O点的水平线.已知一小球从M点出发,初速率为v0,

沿管道MPN运动,到N点的速率为v1,所需时间为t1;若该小球仍由M

点以初速率v0出发,而沿管道MQN运动,到N点的速率为v2,所需时间为t2,则( ) A.v1=v2,t1>t2B.v1t2

C.v1=v2,t1

解析:选A.由于椭圆形管道内壁光滑,小球不受摩擦力作用,因此

小球从M到N的过程机械能守恒,由于M、N在同一高度,根据机械能

守恒定律可知,小球在M、N点的速率相等,选项B、D错误;小球沿

MPN运动的过程中,速率先减小后增大,而沿MQN运动的过程中,速率先增大后减小,两个过程运动的路程相等,到N点速率都为v0,根据速率随时间变化关系图象(如图所示)可知,由于两图象与时间轴所围面积相等,因此t1>t2,选项A正确,C错误.

7.如图甲所示,一个小环套在竖直放置的光滑圆形轨道上做圆周运动.小环从最高点A 滑到最低点B的过程中,其线速度大小的平方v2随下落高度h变化的图象可能是图乙所示四个图中的( )

A .①②

B .③④

C .③

D .④

解析:选A.设小环在A 点的速度为v 0,由机械能守恒定律得-mgh +12m v 2=12

m v 20,得v 2=v 20+2gh ,可见v 2与h 是线性关系,若v 0=0,②正确;若v 0≠0,①正确,故正确选项是

A.

二、多项选择题

8.如图所示装置中,木块与水平桌面间的接触面是光滑的,子弹A

沿水平方向射入木块后留在木块内,将弹簧压缩到最短,则从子弹开始

射入木块到弹簧压缩至最短的整个过程中( )

A .子弹与木块组成的系统机械能守恒

B .子弹与木块组成的系统机械能不守恒

C .子弹、木块和弹簧组成的系统机械能守恒

D .子弹、木块和弹簧组成的系统机械能不守恒

解析:选BD.从子弹射入木块到木块将弹簧压缩至最短的整个过程中,由于存在机械能与内能的相互转化,所以对整个系统机械能不守恒.对子弹和木块,除摩擦生热外,还要克服弹簧弹力做功,故机械能也不守恒.

9.如图所示,在两个质量分别为m 和2m 的小球a 和b 之间,

用一根轻质细杆连接,两小球可绕过轻杆中心的水平轴无摩擦转动,

现让轻杆处于水平位置,静止释放小球后,重球b 向下转动,轻球a 向上转动,在转过90°的过程中,以下说法正确的是( )

A .b 球的重力势能减少,动能增加

B .a 球的重力势能增加,动能减少

C .a 球和b 球的机械能总和保持不变

D .a 球和b 球的机械能总和不断减小

解析:选AC.在b 球向下、a 球向上摆动过程中,两球均在加速转动,两球动能增加,同时b 球重力势能减少,a 球重力势能增加,A 正确,B 错误;a 、b 两球组成的系统只有重力和系统内弹力做功,系统机械能守恒,C 正确,D 错误.

10.两个质量不同的小铁块A 和B ,分别从高度相同的都是光滑的

斜面和圆弧面的顶点滑向底部,如图所示.如果它们的初速度都为0,

则下列说法正确的是( )

A .下滑过程中重力所做的功相等

B .它们到达底部时动能相等

C .它们到达底部时速率相等

D .它们在最高点时的机械能和它们到达最低点时的机械能大小各自相等

解析:选CD.小铁块A 和B 在下滑过程中,只有重力做功,机械能守恒,则由mgH =12

m v 2,得v =2gH ,所以A 和B 到达底部时速率相等,故C 、D 正确;由于A 和B 的质量不同,所以下滑过程中重力所做的功不相等,到达底部时的动能也不相等,故A 、B 错误.

三、非选择题

11.如图所示,位于竖直平面内的光滑轨道,由一段倾斜的

直轨道和与之相切的圆形轨道连接而成,圆形轨道的半径为R .一

质量为m 的小物块从斜轨道上某处由静止开始下滑,然后沿圆形

轨道运动.要求小物块能通过圆形轨道的最高点,且在该最高点与轨道间的压力不能超过5mg (g 为重力加速度).求小物块初始位置相对于圆形轨道底部的高度h 的取值范围.

解析:设小物块在圆形轨道最高点的速度为v ,由机械能守恒定律得

mgh =2mgR +12

m v 2 ① 小物块在最高点受到的重力与弹力的合力提供向心力,有mg +F N =m v 2R

② 小物块能通过最高点的条件是F N ≥0

③ 由②③式得v ≥gR

④ 由①④式得h ≥52

R ⑤ 按题目要求:F N ≤5mg ,由②式得v ≤6gR

由①⑥式得h ≤5R

所以高度h 的取值范围是52

R ≤h ≤5R . 答案:52

R ≤h ≤5R 12.如图所示,轻弹簧一端与墙相连处于自然状态,质量为4 kg 的

木块沿光滑的水平面以5 m/s 的速度运动并开始挤压弹簧,求:

(1)弹簧的最大弹性势能;

(2)木块被弹回速度增大到3 m/s 时弹簧的弹性势能.

解析:(1)木块压缩弹簧的过程中,木块和弹簧组成的系统机械能守恒,弹性势能最大时,对应木块的动能为零,故有:

E pm =12m v 20=12

×4×52 J =50 J . (2)由机械能守恒有12m v 20=E p1+12m v 21

12×4×52 J =E p1+12

×4×32 J 得E p1=32 J .

答案:(1)50 J (2)32 J

(完整版)提公因式法因式分解练习题

因式分解---------提公因式法 下列从左到右的变形中,哪些是因式分解,哪些不是。 (1))2(3362 2 3 b a a b a a -=- (2))1(2 3 2 x x x x --=+- (3)))((2 2 b ab a b a ++-33b a -= (4))3)(2(--x x 652+-=x x (5)㎡=m ×m (6)㎡+m=m 3( ) 二、用提公因式法因式分解(一) (1)332168b a ab - (2)22mn n m +- (3)2 515x xy -- (4)3224 1ab b a - (5)ab b a b a -+2233 (6) 3 22316128ay y a y a -+- (7)am m a m a 126323+--(8)xy y x y x ++-2 2 3 2 用提公因式法因式分解(二) (1)2 )()(b a b a +-+ (2))()(x y y y x x -+- (3))(2)(62 n m n m +-+(4))(2)(32 y x x y -+- (5))()(3y x x y x ----(6)2 2 )()(m n n n m m --- (7))(4)(6p q q q p p +-+ (8))(4)(122 x y ab y x b a --- (9)))(())((y x b a y x b a -+-++ 用提公因式法因式分解(三) (1))(2)(72a b y b a x --- (2) )3()3(52 2x a x --- (3) 23)()(2b a b a +-+ (4)2 22)3()3(a b x b a x --- 5))(3)(2p q b q p a ---(6)2 2 3 )1(8)1(6x p x p --- (7)2 )1()1(---a a a (8)2 2 )()()(b a b a b a --+- (9))1()1(2)1(3x c x b x a -+---- (10))32()23()1(2x x x -+-- 用提公因式法因式分解(四) (1)2 )())((y x x y x y x x +--+

初三英语寒假培优课试题

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因式分解提公因式法含答案

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4.多项式-7ab+14abx-49aby的公因式是________. 5.3x2y3,2x2y,-5x3y2z的公因式是________. 6.下列各式用提公因式法分解因式,其中正确的是(). A.5a3+4a2-a=a(5a2+4a) B.p(a-b)2+pq(b-a)2=p(a-b)2(1+q) C.-6x2(y-z)3+x(z-y)3=-3x(z-y)2(2x-z+y) D.-x n-x n+1-x n+2=-x n(1-x+x2) 7.把多项式a2(x-2)+a(2-x)分解因式等于(). A.(x-2)(a2+a) B.(x-2)(a2-a) C.a(x-2)(a-1) D.a(x-2)(a+1) 8.下列变形错误的是(). A.(y-x)2=(x-y)2 B.-a-b=-(a+b) C.(a-b)3=-(b-a)3 D.-m+n=-(m+n)

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5.Where does this conversation probably take place? A .At home .B.In a stadium.C.In the hospital . 6.Which dress does the man like? A .The red one.B.The white one.C.The purple one. 7.How many desks are there for the parents? A .Sixteen.B.Twenty-five .C.Twenty-six . 8.Who is the man speaking to? A .Mrs .Green.B.Dr.Brown .C.Joe. 9.Which bus does the man want to take? A .A number 5.B.A number 3 .C.A number 63. 10.How is the woman going to the airport? A .By taxi .B.By train .C.By bus. B) 听下面 3 段对话或独白.每段对话或独白后有几个小题,从题中所给的A、B、C 三个 选项中选出最佳选项.听每段对话或独白前,你将有时间阅读各个小题,每小题 5 秒钟;听完后,各小题将给出 5 秒钟的答题时间.每段对话或独向读两遍. 听第1段材料,回答11—12小题. 11.What are they talking about? A .Films .B.Stories.C.Teenagers. 12.Why does Sandy think fights in actions films are terrible? A .Because they are really exciting . B.Because they have happy endings. C.Because they have a bad effect on people. 听第二段对话,回答13—15小题. 13.Who has a problem? A .Mary ’s mother.B.Mr Li .C.Mary . 14.How does Mr Li think an MP4 player to Mary? A .Useless.B.Necessary.C.Important . 15.What will her mother do if Mary says sorry to her mother? A .She will be angry .B.She will be pleased.C.She will buy her an MP4 player . 听下面一段独白,回答16—20小题. 16.What can we learn from the passage? A .Some social customs(风俗)in Britain . B.Some social customs in Canada. C.The weather in Britain . 17.What are the British happy to talk about? A .Money .B.Age.C.Weather. 18.What should a man do when he is walking into a room with a woman? A .Ask about her age. B.Follow her .

(完整版)因式分解练习题(提取公因式)

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因式分解专题1_用提公因式法(含答案)

1、用提公因式法把多项式进行因式分解 【知识精读】 如果多项式的各项有公因式,根据乘法分配律的逆运算,可以把这个公因式提到括号外面,将多项式写成因式乘积的形式。 提公因式法是因式分解的最基本也是最常用的方法。它的理论依据就是乘法分配律。多项式的公因式的确定方法是: (1)当多项式有相同字母时,取相同字母的最低次幂。 (2)系数和各项系数的最大公约数,公因式可以是数、单项式,也可以是多项式。 下面我们通过例题进一步学习用提公因式法因式分解 【分类解析】 1. 把下列各式因式分解 (1)-+--+++a x abx acx ax m m m m 2213 (2)a a b a b a ab b a ()()()-+---32222 分析:(1)若多项式的第一项系数是负数,一般要提出“-”号,使括号内的第一项系数是正数,在提出“-”号后,多项式的各项都要变号。 解:-+--=--+++++a x abx acx ax ax ax bx c x m m m m m 221323() (2)有时将因式经过符号变换或将字母重新排列后可化为公因式,如:当n 为自然数时,() ()()()a b b a a b b a n n n n -=--=----222121;,是在因式分解过程中常用的因式 变换。 解:a a b a b a ab b a ()()()-+---32222 ) 243)((]2)(2))[(() (2)(2)(222223b b ab a b a a b b a a b a b a a b a ab b a a b a a ++--=+-+--=-+-+-= 2. 利用提公因式法简化计算过程 例:计算1368 987521136898745613689872681368987123?+?+?+?

九年级英语期末试卷

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《因式分解-提公因式法》知识点归纳

《因式分解-提公因式法》知识点归纳★★ 知识体系梳理 ◆ 因式分解------把一个多项式变成几个整式的积的形式;(化和为积) 注意: 、因式分解对象是多项式; 2、因式分解必须进行到每一个多项式因式不能再分解为止; 3、可运用因式分解与整式乘法的互逆关系检验因式分解的正确性; ◆ 分解因式的作用 分解因式是一种重要的代数恒等变形,它有着广泛的应用,常见的用途有化简多项式和进行简便运算,恰当的运用分解因式,常可以使计算化繁为简。 ◆ 分解因式的一些原则 (1)提公因式优先的原则.即一个多项式的各项若有公因式,分解时应首先提取公因式。 (2)分解彻底的原则.即分解因式必须进行到每一个

多项式因式都再不能分解为止。 (3)首项为负的添括号原则.即如果多项式的首项系数为负,应先添上带“-”号的括号,并遵循添括号法则。 ◆ 因式分解的首要方法—提公因式法 、公因式:一个多项式每项都含有的公共的因式,叫做这个多项式各项的公因式。 2、提公因式法:如果一个多项式的各项含有公因式,可以逆用乘法分配律,把各项共有的 因式提出以分解因式的方法,叫做提公因式法。 3、使用提取公因式法应注意几点: (1)提取的“公因式”可以是数、单项式,也可以是一个多项式,是一个整体。 (2)公因式必须是多项式的每一项都有的因式,在提取公因式时,要把这些公共的因式全部找出来,并提到括号外面去,才算完成了提取公因式。(找最高公因式)(3)对多项式中的每一项的数字系数,在提取时要提出这些数字系数的最大公约数,各项都含有相同的字母,要提取相同字母的指数的最低指数。 ◆ 提公因式法分解因式的关键: 、确定最高公因式;(各项系数的最大公约数与相同因

因式分解一_提取公因式法和公式法_超经典

因式分解(一) ——提取公因式与运用公式法 【学习目标】(1)让学生了解什么是因式分解; (2)因式分解与整式的区别; (3)提公因式与公式法的技巧。 【知识要点】 1、提取公因式:型如()ma mb mc m a b c ++=++,把多项式中的公共部分提取出来。 ☆提公因式分解因式要特别注意: (1)如果多项式的首项系数是负的,提公因式时要将负号提出,使括号内第一项的系数是正的, 并且注意括号内其它各项要变号。 (2)如果公因式是多项式时,只要把这个多项式整体看成一个字母,按照提字母公因式的办法提出。 (3)有时要对多项式的项进行适当的恒等变形之后(如将a+b-c 变成-(c-a-b )才能提公因式, 这时要特别注意各项的符号)。 (4)提公因式后,剩下的另一因式须加以整理,不能在括号中还含有括号,并且有公因式的还应继续提。 (5)分解因式时,单项式因式应写在多项式因式的前面。 2、运用公式法:把我们学过的几个乘法公式反过来写就变成了因式分解的形式: ()()22a b a b a b -=+-; ()2 222a ab b a b ±+=±。 平方差公式的特点是:(1) 左侧为两项;(2) 两项都是平方项;(3) 两项的符号相反。 完全平方公式特点是: (1) 左侧为三项;(2) 首、末两项是平方项,并且首末两项的符号相同; (3) 中间项是首末两项的底数的积的2倍。 ☆运用公式法分解因式,需要掌握下列要领: (1)我们学过的三个乘法公式都可用于因式分解。具体使用时可先判断能否用公式分解,然后再选择适当公式。(2)各个乘法公式中的字母可以是数,单项式或多项式。 (3)具体操作时,应先考虑是否可提公因式,有公因式的要先提公因式再运用公式。 (4)因式分解一定要分解到不能继续分解为止,分解之后一定要将同类项合并。 【经典例题】 例1、找出下列中的公因式: (1) a 2b ,5ab ,9b 的公因式 。 (2) -5a 2,10ab ,15ac 的公因式 。 (3) x 2y(x -y),2xy(y -x) 的公因式 。

九年级英语期末测试卷及答案(一)

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