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课后强化训练42 综合型问题

课后强化训练42 综合型问题
课后强化训练42 综合型问题

课后强化训练42 综合型问题

一、选择题

(第1题)

1.如图,A ,B 是半径为1的⊙O 上两点,且OA ⊥OB .点P 从点A 出发,在⊙O 上以每秒一个单位长度的速度匀速运动,回到点A 运动结束. 设运动时间为x ,弦BP 的长度为y ,那么下列图象中可能表示y 与x 的函数关系的是(D )

A. ① B .④

C .②或④

D .①或③

【解析】 当点P 逆时针运动时,从点A 到点B ,弦BP 逐渐变短,从点B 到点B 关于点O 的对称点B ′,弦BP 逐渐变长,到点B ′时,达到最大值,从点B ′到点A ,又逐渐变短;当点P 顺时针运动时,从点A 到点B ′,弦BP 逐渐变长,到点B ′时,达到最大值,从点B ′到点B ,逐渐变短至0,从点B 到点A ,又逐渐变长.

(第2题)

2.如图,AB 是⊙O 的直径,CD ,EF 是⊙O 的弦,且AB ∥CD ∥EF .若AB =10,CD =6,EF =8,则图中阴影部分的面积是(A )

A.25

2π B .10π C .24+4π D .24+5π

(第2题解)

【解析】 如解图,作直径CG ,连结OD ,OE ,OF ,DG . ∵CG 是⊙O 的直径, ∴∠CDG =90°, ∴DG =

CG 2-CD 2=

102-62=8.

又∵EF =8,∴DG =EF , ∴DG ︵=EF ︵

,∴S 扇形ODG =S 扇形OEF .

∵AB ∥CD ∥EF ,∴S △OCD =S △ACD ,S △OEF =S △AEF ,

∴S 阴影=S 扇形OCD +S 扇形OEF =S 扇形OCD +S 扇形ODG =S 半圆=12π×52=25

2

π.

二、填空题 3.我国三国时期数学家赵爽为了证明勾股定理,创制了一幅“弦图”,后人称其为“赵爽弦图”,如图①所示.在图②中,若正方形ABCD 的边长为14,正方形IJKL 的边长为2,且IJ ∥AB ,则正方形EFGH 的边长为__10__.

,(第3题))

【解析】 易得正方形ABCD 是由八个全等的直角三角形和一个小正方形组成的. 设EJ =x ,则HJ =x +2,

∴S 正方形ABCD =8×1

2x (x +2)+22=142,

化简,得x 2+2x -48=0,

解得x 1=6,x 2=-8(不合题意,舍去). ∴正方形EFGH 的边长为

62+82=10.

(第4题)

4.如图,一次函数y =-x +b 与反比例函数y =4

x (x >0)的图象交于A ,B 两点,与x 轴,

y 轴分别交于C ,D 两点,连结OA ,OB ,过点A 作AE ⊥x 轴于点E ,交OB 于点F ,设点A 的横坐标为m .

(1)b =4

m

+m (用含m 的代数式表示).

(2)若S △OAF +S 四边形EFBC =4,则m 的值是__2__. 【解析】 (1)把x =m 代入y =4x ,得y =4

m ,

∴点A 的坐标为?

???m ,4

m . 把点A 的坐标代入y =-x +b ,得4

m =-m +b ,

解得b =4

m

+m .

(2)过点B 作x 轴的垂线,垂足为G .

将y =-x +b 代入y =4x ,得-x +b =4

x ,化简,得x 2-bx +4=0,∴x A ·x B =4.

又∵x A =m ,∴x B =4m ,∴y B =4

x B =m ,

∴点B 的坐标为????4

m ,m , ∴直线OB 的函数表达式为y =m 2

4x .

将x =m 代入y =m 24x ,得y =m 3

4,

∴点F 的坐标为????m ,m 3

4. ∵S △OAE =S △OBG =1

2

×4=2,

∴S △OAE +S △OBG =4=S △OAF +S 四边形EFBC , ∴S △BCG =2S △OEF ,即12CG ·BG =2×1

2

OE ×EF ,

∴12m ·m =2×12m ×m 3

4

,解得m 1=2,m 2=-2(不合题意,舍去).

∴m 的值为 2.

(第5题)

5.如图,有一个不确定的正方形ABCD ,它的两个相对的顶点A ,C 分别在边长为1的正六边形的一组对边上,另外两个顶点B ,D 在正六边形内部(包括边界),则正方形边长a 的取值范围是

6

2

≤a ≤3-3.

导学号:71594049

【解析】 ∵AC 为对角线,故当AC 最小时,正方形的边长最小.

①当点A ,C 都在对边的中点时,如解图中的正方形ABCD ,此时AC 取得最小值.

,(第5题解))

∵正六边形的边长为1,∴AC =3, ∴a 2+a 2=AC 2=(3)2,∴a =

32=62

. ②当正方形的四个顶点都在正六边形的边上时,如解图中的正方形A ′B ′C ′D ′,此时AC 取得最大值.

建立如解图所示的坐标系,连结OA ′,OB ′, 设点A ′?

??

?t ,

32.

∵OA ′=OB ′,OB ′⊥OA ′,∴点B ′?

??

?-

32, t . 设直线MN 的函数表达式为y =kx +b ,把点M (-1,0),N ????-12,-3

2的坐标代入,

?????-k +b =0,-12k +b =-32,

解得???

??k =-3,

b =- 3. ∴直线MN 的函数表达式为y =-3x - 3. 把点B ′?

???-

32, t 的坐标代入y =-3x -3,得t =32-3,∴点A ′????32

-3,

32,B ′???

?-32,32-3,

∴a =A ′B ′=

????32-3+322+???

?32-32+32

=3- 3.

∴a 的取值范围是6

2

≤a ≤3- 3. 三、解答题

(第6题)

6.如图,矩形AOCB 的顶点A ,C 分别位于x 轴和y 轴的正半轴上,线段OA ,OC 的长度满足方程|x -15|+y -13=0(OA >OC ),直线y =kx +b 分别与x 轴、y 轴相交于M ,N

两点,将△BCN 沿直线BN 折叠,点C 恰好落在直线MN 上的点D 处,且tan ∠CBD =3

4.

(1)求点B 的坐标.

(2)求直线BN 的函数表达式.

(3)将直线BN 以每秒1个单位长度的速度沿y 轴向下平移,求直线BN 扫过矩形AOCB 的面积S 关于运动时间t (0<t ≤13)的函数表达式.

【解析】 (1)∵|x -15|+y -13=0,∴x =15,y =13, ∴OA =BC =15,AB =OC =13,∴点B (15,13).

(第6题解①)

(2)如解图①,过点D 作DE ⊥OA 于点E ,延长ED 交CB 于点F . 由折叠的性质可知,BD =BC =15,∠BDN =∠BCN =90°. ∵tan ∠CBD =34,∴DF BF =3

4

,且BF 2+DF 2=BD 2=152,

∴BF =12,DF =9,

∴OE =CF =BC -BF =15-12=3,DE =EF -DF =13-9=4.

∵∠CND +∠CBD =360°-90°-90°=180°,且∠ONM +∠CND =180°, ∴∠ONM =∠CBD ,∴△BDF ∽△NMO ,∴OM ON =FD FB =3

4.

易知DE ∥ON ,∴EM ED =OM ON =3

4,

OM -34=3

4

,解得OM =6, ∴ON =8,即点N (0,8).

把点N ,B 的坐标代入y =kx +b ,得?????b =8,

15k +b =13,解得??

???k =1

3,b =8,

∴直线BN 的函数表达式为y =1

3

x +8.

(3)设直线BN 平移后交y 轴于点N ′,交AB 于点B ′. 当点N ′在ON 上,即0<t ≤8时,如解图②.

根据题意可知四边形BNN ′B ′为平行四边形,且NN ′=t ,∴S =NN ′·OA =15t

.

(第6题解)

当点N ′在y 轴的负半轴上,即8<t ≤13时,设直线B ′N ′交x 轴于点G ,如解图③. ∵NN ′=t ,∴可设直线B ′N ′的函数表达式为y =1

3x +8-t .

令y =0,得x =3t -24,∴OG =3t -24.

∵ON =8,NN ′=t ,∴ON ′=t -8,

∴S =S ?BNN ′B ′-S △OGN ′=15t -12(t -8)(3t -24)=-3

2t 2+39t -96.

综上所述,S 关于t 的函数表达式为

S =?????15t (0<t ≤8),

-32t 2

+39t -96(8<t ≤13).

7.如图,在平面直角坐标系xOy 中,点A 的坐标为(0,1),取一点B (b ,0),连结AB ,

作线段AB 的垂直平分线l 1,过点B 作x 轴的垂线l 2,记l 1,l 2的交点为P .

(1)当b =3时,在图中补全图形(尺规作图,不写作法,保留作图痕迹). (2)小慧多次取不同数值b ,得出相应的点P ,并把这些点用平滑的曲线连结起来,发现:这些点P 竟然在一条曲线l 上!

①设点P 的坐标为(x ,y ),试求y 与x 之间的函数表达式,并指出曲线l 是哪种曲线. ②设点P 到x 轴,y 轴的距离分别为d 1,d 2,求d 1+d 2的范围;当d 1+d 2=8时,求点P 的坐标.

③将曲线l 在直线y =2下方的部分沿直线y =2向上翻折,得到一条“W”形状的新曲线,若直线y =kx +3与这条“W”形状的新曲线有4个交点,直接写出k 的取值范围.

,(第7题))

【解析】 (1)如解图①.

(第7题解①)

(2)①当x >0时,如解图①,连结AP ,过点P 作PE ⊥y 轴于点E . ∵l 1垂直平分AB , ∴P A =PB =y .

在Rt △APE 中,EP =OB =x ,AE =OE -OA =y -1. 由勾股定理,得(y -1)2+x 2=y 2. 整理,得y =12x 2+1

2

.

当x ≤0时,点P (x ,y )同样满足y =12x 2+1

2

.

∴曲线l 就是二次函数y =12x 2+1

2的图象,即曲线l 是一条抛物线.

②由题意可知,d 1=12x 2+1

2,d 2=|x |.

∴d 1+d 2=12x 2+1

2

+|x |.

当x =0时,d 1+d 2有最小值1

2,

∴d 1+d 2的范围是d 1+d 2≥1

2.

当d 1+d 2=8时,12x 2+1

2

+|x |=8.

当x ≥0时,原方程可化为12x 2+1

2+x =8,

解得 x 1=3,x 2=-5(不合题意,舍去). 当x <0时,原方程可化为12x 2+1

2-x =8,

解得x 1=-3,x 2=5(不合题意,舍去). 把x =±3代入y =12x 2+1

2,得 y =5.

∴点P 的坐标为(3,5)或(-3,5). ③k 的取值范围是-33<k <3

3

.

解答过程如下:如解图②.

(第7题解②)

把y =2代入y =12x 2+1

2

,得x 1=-3,x 2= 3.

∴直线y =2与抛物线y =12x 2+1

2的两个交点的坐标为(-3,2)和(3,2).

当直线y =kx +3过点(-3,2)时,可求得 k =3

3; 当直线y =kx +3过点(3,2)时,可求得 k =-

33

. 故当直线y =kx +3与这条“W ”形状的新曲线有4个交点时,k 的取值范围是-3

3

<k <

33

.

(第8题)

8.如图,在Rt △ABC 中,∠C =90°,AC =8,BC =6,点P 以每秒1个单位长度的速度从点A 向点C 运动,同时点Q 以每秒2个单位长度的速度沿A →B →C 方向运动,它们到点C 后都停止运动,设点P ,Q 运动的时间为t (s).

(1)在运动过程中,求P ,Q 两点之间距离的最大值.

(2)P ,Q 两点在运动过程中,是否存在时间t ,使得△PQC 为等腰三角形?若存在,求出此时的t 值;若不存在,请说明理由.

【解析】 (1)如解图①,过点Q 作QE ⊥AC 于点E ,连结PQ . ∵∠C =90°,∴QE ∥BC , ∴△ABC ∽△AQE , ∴AQ AB =AE AC =QE BC

. ∵AQ =2t ,AP =t ,∠C =90°,AC =8,BC =6,

∴AB =10,∴2t 10=t +PE 8=QE

6,

∴PE =35t ,QE =6

5t ,

∴PQ 2=QE 2+PE 2, ∴PQ =355

t .

易知当点Q 与点B 重合时,PQ 的值最大, ∴当t =5时,PQ 的最大值=3 5.

,(第8题解))

(2)存在.

如解图②,连结CQ ,PQ ,过点Q 作QE ⊥AC 于点E ,则QE =6

5t ,CE =AC -AE =8

-85t ,PQ =355

t , ∴CQ =

QE 2

+CE 2

????65t 2+???

?8-85t 2

4t 2-128

5t +64=2

t 2-32

5

t +16.

①当CQ =CP ,即2

t 2-325t +16=8-t 时,t 1=16

5

,t 2=0(不合题意,舍去).

②当PQ =CQ ,即35

5

t =2

t 2-325t +16时,t 1=40

11

,t 2=8(不合题意,舍去).

③当PQ =PC ,即35

5

t =8-t 时,t =65-10.

综上所述,当t =165或40

11

或65-10时,△PQC 为等腰三角形.

(第9题)

9.如图,在平面直角坐标系xOy 中,将二次函数y =x 2-1的图象M 沿x 轴翻折,把所得到的图象向右平移2个单位长度后再向上平移8个单位长度,得到二次函数的图象N .

(1)求N的函数表达式.

(2)设点P(m,n)是以点C(1,4)为圆心,1为半径的圆上一动点,二次函数的图象M与x轴相交于A,B两点,求P A2+PB2的最大值.

(3)若一个点的横坐标与纵坐标均为整数,则该点称为整点.求M与N所围成的封闭图形内(包括边界)整点的个数.

【解析】(1)二次函数y=x2-1的图象M沿x轴翻折得到函数的表达式为y=-x2+1,此时顶点坐标为(0,1).

将此图象向右平移2个单位长度后再向上平移8个单位长度,得到二次函数图象N的顶点坐标为(2,9),

故N的函数表达式为y=-(x-2)2+9=-x2+4x+5.

(第9题解①)

(2)易得点A(-1,0),B(1,0),

∴P A2+PB2=(m+1)2+n2+(m-1)2+n2=2(m2+n2)+2=2PO2+2,

∴当PO最大时,P A2+PB2最大.

如解图①,延长OC与⊙O相交于点P,此时PO最大.

∵点C(1,4),∴OC=42+1=17,∴PO的最大值为OC+PC=17+1,

∴P A2+PB2的最大值为2(17+1)2+2=38+417.

(3)M与N所围成的封闭图形如解图②所示.

,(第9题解②))

由图象可知,M与N所围成的封闭图形内(包括边界)整点的个数为25.

10.如图,抛物线l :y =-1

2(x -t )(x -t +4)(常数t >0)与x 轴从左到右的交点为B ,A ,

过线段OA 的中点M 作MP ⊥x 轴,交反比例函数y =k

x (k >0,x >0)的图象于点P ,且OA ·MP

=12.

,(第10题))

(1)求k 的值.

(2)当t =1时,求AB 的长,并求直线MP 与抛物线l 的对称轴之间的距离.

(3)把抛物线l 在直线MP 左侧部分的图象(含与直线MP 的交点)记为G ,用t 表示图象G 最高点的坐标.

(4)设抛物线l 与反比例函数有个交点的横坐标为x 0,且满足4≤x 0≤6,通过抛物线l 的位置随t 变化的过程,直接写出t 的取值范围.

【解析】 (1)设点P (x ,y ),则MP =y . 由OA 的中点为M ,知OA =2x .

代入OA ·MP =12,得2x ·y =12,即xy =6. ∴k =xy =6.

(2)当t =1时,y =-12(x -1)(x +3),令y =0,则0=-1

2(x -1)(x +3),解得x 1=1,x 2=

-3.

∴点B (-3,0),A (1,0),∴AB =4.

易知抛物线l 的对称轴为直线x =-1,点M ????

12,0, ∴MP 与抛物线l 的对称轴之间的距离为12-(-1)=32.

(3)∵点A (t ,0),B (t -4,0), ∴抛物线l 的对称轴为直线x =t -2. 又∵直线MP 为x =t

2

∴当t -2≤t

2,即t ≤4时,顶点(t -2,2)就是图象G 的最高点;

当t >4时,抛物线l 与MP 的交点???

?t 2,-1

8t 2+t 就是图象G 的最高点. (4)对于y =6x ,当4≤x 0≤6时,1≤y 0≤3

2,即抛物线l 与反比例函数在点C ????4,32,D (6,1)之间的一段有个交点.

①由32=-1

2(4-t )(4-t +4),得t 1=5,t 2=7;

②由1=-1

2

(6-t )(6-t +4),得t 3=8-2,t 4=8+ 2.

随着t 的逐渐增大,抛物线l 的位置随点A (t ,0)向右平移,如解图所示.

(第10题解)

当t =5时,抛物线l 的右侧过点C ;当t =8-2<7时,抛物线l 的右侧过点D ,故5≤t ≤8-2符合题意.

当8-2<t <7时,抛物线l 的右侧离开了点D ,而左侧未到点C ,即抛物线l 与该段无交点,故舍去.

当t =7时,抛物线l 的左侧过点C ;当t =8+2时,抛物线l 的左侧过点D ,故7≤t ≤8+2符合题意.

综上所述,5≤t ≤8-2或7≤t ≤8+ 2.

11.小明是一个喜欢探究钻研的学生,他在和同学们一起研究某条抛物线y =ax 2(a <0)的性质时,将一把直角三角尺的直角顶点置于平面直角坐标系的原点O ,两直角边与该抛物线交于A ,B 两点,请解答以下问题:

(1)若测得OA =OB =22(如图①),求a 的值. (2)对同一条抛物线,小明将三角尺绕点O 旋转到如图②所示的位置时,过点B 作BF ⊥x

轴于点F ,测得OF =1,写出此时点B 的坐标,并求点A 的横坐标.

(3)对该抛物线,小明将三角尺绕点O 旋转任意角度时惊奇地发现,交点A ,B 的连线段总经过一个固定的点,试说明理由并求出该点的坐标.

导学号:71594050

,(第11题))

【解析】 (1)设线段AB 与y 轴的交点为C ,由抛物线的对称性可得C 为AB 的中点. ∵OA =OB =22,∠AOB =90°, ∴AC =OC =BC =2,∴点B (2,-2).

(第11题解)

把点B (2,-2)的坐标代入y =ax 2(a <0),得a =-1

2.

(2)如解图,过点A 作AE ⊥x 轴于点E . ∵点B 的横坐标为1, ∴点B ????1,-12, ∴BF =1

2

.

∵∠AOB =90°,∠OFB =90°,∴∠AOE =∠OBF . 又∵∠AEO =∠OFB =90°,

∴△AEO ∽△OFB ,∴AE OE =OF BF =1

1

2=2,

∴AE =2OE .

设点A ????-m ,-12m 2(m >0),则OE =m, AE =12m 2, ∴1

2m 2=2m ,∴m =4,即点A 的横坐标为-4.

(3)设点A ????-m ,-12m 2(m >0),B ?

???n ,-1

2n 2(n >0),直线AB 的函数表达式为y =kx +b , 则???-mk +b =-1

2m 2 ①,

nk +b =-1

2n 2

②,

①×n +②×m ,得(m +n )b =-12(m 2n +mn 2)=-12mn (m +n ),∴b =-1

2mn .

由题意可知,△AEO ∽△OFB ,∴AE OF =OE

BF ,

∴12m 2n =m 12

n 2,∴mn =4,∴b =-12

×4=-2.

∴不论k 为何值,直线AB 恒过点(0,-2). 12.问题背景:

如图①,在四边形ADBC 中,∠ACB =∠ADB =90°,AD =BD ,探究线段AC ,BC ,CD 之间的数量关系.

小吴同学探究此问题的思路是:将△BCD 绕点D 逆时针旋转90°到△AED 处,点B ,C 分别落在点A ,E 处(如图42-6②),易证点C ,A ,E 在同一条直线上,并且△CDE 是等腰直角三角形,故CE =2CD ,从而得出结论:AC +BC =2CD .

简单应用:

(1)在图①中,若AC =2,BC =22,则CD =__3__.

(2)如图③,AB 是⊙O 的直径,点C ,D 在⊙O 上,AD ︵=BD ︵

,若AB =13,BC =12,求CD 的长.

,(第12题))

拓展规律:

(3)如图④,∠ACB =∠ADB =90°,AD =BD ,若AC =m ,BC =n (m

(4)如图⑤,∠ACB =90°,AC =BC ,P 为AB 的中点,若点E 满足AE =1

3AC ,CE =CA ,

Q 为AE 的中点,则线段PQ 与AC 的数量关系是2PQ =1+356AC 或2PQ =35-1

6

AC .

导学号:71594051

【解析】 (1)由题意,得AC +BC =2CD , ∴2+22=2CD ,∴CD =3.

(第12题解①)

(2)如解图①,连结AC ,BD ,AD . ∵AB 是⊙O 的直径, ∴∠ADB =∠ACB =90°. ∵AD ︵=BD ︵

,∴AD =BD .

将△BCD 绕点D 顺时针旋转90°到△AED 处, 则∠DAE =∠DBC ,△EDC 是等腰直角三角形. ∵∠DBC +∠DAC =180°,∴∠EAD +∠DAC =180°, ∴E ,A ,C 三点共线.

∵AB =13,BC =12,∴由勾股定理,得AC =5. ∵BC =AE ,∴CE =AE +AC =17. ∵△EDC 是等腰直角三角形, ∴CE =2CD ,∴CD =1722

.

(3)以AB 为直径作⊙O ,连结DO 并延长,交⊙O 于点D 1,连结D 1A ,D 1B ,D 1C ,如

解图②.

由(2)可知,AC +BC =2D 1C ,∴D 1C =2(m +n )

2.

又∵D 1D 是⊙O 的直径,∴∠DCD 1=90°.

∵AC =m ,BC =n ,∴由勾股定理,得AB 2=m 2+n 2, ∴D 1D 2=AB 2=m 2+n 2. ∵D 1C 2+CD 2=D 1D 2,

∴CD 2

=m 2

+n 2

-(m +n )22=(m -n )22

.

∵m

2

.

,(第12题解))

(4)①当点E 在直线AC 的左侧时,如解图③,连结CQ ,CP . ∵AC =BC ,∠ACB =90°,P 是AB 的中点, ∴AP =CP ,∠APC =90°.

又∵CA =CE ,Q 是AE 的中点,∴∠CQA =90°. 设AC =a ,则AE =13a ,AQ =12AE =16a .

在Rt △ACQ 中,由勾股定理,得CQ =AC 2-AQ 2=

35

6

a . 由(2)可知,AQ +CQ =2PQ ,

∴2PQ =16a +35

6a ,∴2PQ =1+356

AC .

(第12题解④)

②当点E 在直线AC 的右侧时,如解图④,连结CQ ,CP . 设AC =a ,同①可知,AQ =16a ,CQ =35

6a ,

由(3)的结论可知,PQ =2

2

(CQ -AQ ), ∴2PQ =35-1

6

AC .

综上所述,线段PQ 与AC 的数量关系是2PQ =1+356AC 或2PQ =35-1

6AC .

研究生英语综合教程(课后习题答案)

Unit One Task 1 1.A 2.C 3.B 4.C 5.D 6.D 7.D 8.C 9.A 10.D 11.A 12.B Task 2 1.public(c) 2.discipline(b) 3.strength(a) 4.reference(a) 5.strength(d) 6.public(a) 7.demonstrated(b) 8.discipline(c) 9.references(c) 10.personality(a) 11.discipllining(d) 12.demonstrates(a) 13.public(d) 14.reference(b) 15.personality(c) Task 3 1.employment 2.paid 3.adjust 4.setting 5.discouraged 6.credit 7.cite 8.demonstrate 9.teamwork 10.rules Unit Two Task 1 1.A 2.B 3.B 4.C 5.B 6.A 7.B 8.C 9.A 10.C Task 2 1. bud (n.); budding (adj.) 2. access (n.); access (v.) 3. taste (n.);tasted (v.) 4. fool (n.); fooling (v.) 5. produces (v.); produce (n.) 6. garnish (v.); garnishes (n.) 7. reigns (v.); reign (n.) 8. concern (n.); concerned (v.) 9. named (v.); name (n.) 10. practiced (v.); practice (n.) Task 3 1) integration 2) choice 3) handed 4) aspiring 5) steaming 6) masterpieces 7) pleasure 8) partake 9) amazing 10) presented Unit Three Task 1 1.A 2.B 3.C 4.B 5.A 6.B 7.C 8.A Task 2 1. stack up against 2. struck a chord 3. amounted to 4. chopping off 5. appeal to 6. pick up on 7. turned out 8. fade away 9. brought together 10. pulled off 11. thrust upon 12. be kept clear of Task 3 1) swirling 2) delivered 3) glowed 4) intervals 5) converge 6) wanderings 7) navigate 8) jealousy 9) presence 10) absorbed Unit Four Task 1 1.A 2. A 3. C 4. B 5. B 6. C 7. D 8. C 9. A 10. C Task 2 1. maintained (a) 2. romantic (a)

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(完整word版)学术综合英语课后答案解析

Unit 1 C 1.The younger generation should continue to sustain and develop our fine traditions and long-standing culture. 2.In the course of preparing one’s speech, one should be clearly aware of how one could make effective use of statistics and examples to bolster one’s point of view. 3.An impromptu speech is one of the speaking skills that college students should learn and develop through practice. 4.By using simile and metaphor, you can make your language more vivid and more attractive to your audience. 5.The proper examples you cite might help reinforce the impression on your listeners and make your viewpoints more convincing. 6.When you are speaking, you should choose common and easy words and at the same time avoid clutter in your speech.

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李明是学化学的,性格开朗幽默,颇有魅力,但英语成绩不佳,每次只能勉强及格。老师警告他,英语不好会阻碍他拿奖学金,并亮出了自己的王牌:如果李明不努力,就让他考试不过关。老师还告诉他,学习英语不能只为了文凭,否则他即使大学毕业,也还是个半文盲。李明虽然保持镇定,但他明白,他的学业生涯正在攸关之际,必须安心下来埋头学习,坚持不懈。 Li Ming was a chemistry major, a charmer noted for his easygoing and humorous temperament. However, his English was so poor that he always barely got by. The teacher admonished him that his poor English would be an impediment to scholarship. What’s more, she showed her trump card: if Li Ming did not work hard. She would flunk him. He was also told that he should not learn English merely for the sake of his diploma, otherwise, even after graduation from university, he would still be semiliterate. Although Li Ming did not lose his composure, he was well aware that he had to settle down to work and follow through because his academic life was at stake. Unit 2 我的朋友琳达接受过良好的教育,既美丽又端庄,三十好几依然没有人向她求婚。究其原因,她的事业心极强,整日扑在工作上,每天来往于住处和公司之间,根本没有时间和异性交往。一想到女儿这么大了还单身一人,她父母就焦虑不安。他们不知道该如何是好,甚至还去咨询一些社会学专家。但是事情在上个月出现了转机,公司的总部调琳达到培训部。在新的工作岗位上,琳达遇到了第一个触动她心弦的男人。从此,他们几乎每天约会,琳达意识到她会不顾一切地爱这个男人。决定嫁人的时候,她告诉了我这个好消息。虽然琳达的爱情让人想起电影中才会有的浪漫故事,我也担忧未来究竟会怎样,但我还是表达了我由衷的祝福,并爽快答应在婚礼那天做他们的伴娘和伴郎随从中的一员。 Linda, my good friend, has received good education and is both beautiful and elegant. She was not proposed to even when she was well over thirty. The reason is that she, as a career –oriented woman, is devoted to her work. Navigating between home and the company, she had hardly any time to socialize with people of the opposite sex. Her parents were gripped by anxiety at the thought of their daughter still remaining single at such an age. They did not know what to do and even consulted with some sociologists. But the situation began to change last month, when the headquarters of the company transferred Linda to the training department. On the new post, Linda met a man who tugged on her heartstrings for the first time. Ever since then, they dated virtually on a daily basis, and Linda realized that she would love the man beyond all reason. When she decided to take the matrimonial plunge, she informed me. Though Linda’s love is reminiscent of the romance that we see only in movies and I don’t know what the future will hold for her, I give her my heart-felt wishes and agree readily to be a member of the entourage of bridesmaids and groomsmen.

综合英语一课后答案

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基础综合英语_1-5单元课后翻译

作文翻译 Unit 1 李明是学化学的,性格开朗幽默,颇有魅力,但英语成绩不佳,每次只能勉强及格。老师警告他,英语不好会阻碍他拿奖学金,并亮出了自己的王牌:如果李明不努力,就让他考试不过关。老师还告诉他,学习英语不能只为了文凭,否则他即使大学毕业,也还是个半文盲。李明虽然保持镇定,但他明白,他的学业生涯正在攸关之际,必须安心下来埋头学习,坚持不懈。 Li Ming was a chemistry major, a charmer noted for his easygoing and humorous temperament. However, his English was so poor that he always barely got by. The teacher admonished him that his poor English would be an impediment to scholarship. What’s more, she showed her trump card: if Li Ming did not work hard. She would flunk him. He was also told that he should not learn English merely for the sake of his diploma. otherwise, even after graduation from university, he would still be semiliterate. Although Li Ming did not lose his composure, he was well aware that he had to settle down to work and follow through because his academic life was at stake. Unit2 我的朋友琳达接受过良好的教育,既美丽又端庄,三十好几依然没有人向她求婚。究其原因,她的事业心极强,整日扑在工作上,每天来往于住处和公司之间,根本没有时间和异性交往。一想到女儿这么大了还单身一人,她父母就焦虑不安。他们不知道该如何是好,甚至还去咨询一些社会学专家。 但是事情在上个月出现了转机,公司的总部调琳达到培训部。在新的工作岗位上,琳达遇到了第一个触动她心弦的男人。从此,他们几乎每天约会,琳达意识到她会不顾一切地爱这个男人。决定嫁人的时候,她告诉了我这个好消息。 虽然琳达的爱情让人想起电影中才会有的浪漫故事,我也担忧未来究竟会怎样,但我还是表达了我由衷的祝福,并爽快答应在婚礼那天做他们的伴娘和伴郎随从中的一员。 Linda, my good friend, has received good education and is both beautiful and elegant. She was not proposed to even when she was well over thirty. The reason is that she, as a career –oriented woman, is devoted to her work. Navigating between home and the company, she had hardly any time to socialize with people of the opposite sex. Her parents were gripped by anxiety at the thought of their daughter still remaining single at such an age. They did not know what to do and even consulted with some sociologists. But the situation began to change last month, when the headquarters of the company transferred Linda to the training department. On the new post, Linda met a man who tugged on her heartstrings for the first time. Ever since then, they dated virtually on a daily basis, and Linda realized that she would love the man beyond all reason. When she decided to take the matrimonial plunge, she informed me. Though Linda’s love is reminiscent of the romance that we see only in movies and I don’t know what the future will hold for her, I give her my heart-felt wishes and agree readily to be a member of the entourage of bridesmaids and groomsmen. Unit 3 食品供应商缺乏诚信已经成为当今社会的一大问题。部分企业欺骗公众,故意散布假消息,颂扬食品添加剂是食品工业的伟大成就,并声称适量的添加剂对健康有益无害。部分有良知的科学家对食品添加剂的含量和毒性展开了深入的病理学研究。研究结果表明,部分常见的食品添加剂经长期,可能会对健康产生危害,这被认为是食品安全研究方面极为重要的

学术综合英语unit1_5课后习题答案解析

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研究生英语综合教程课后习题答案

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