2018九年級(jí)下數(shù)摸底試一下各的個(gè)項(xiàng)，且有個(gè)項(xiàng)正的，擇確的號(hào)填在題的應(yīng)置上．計(jì)算

(a

)

的結(jié)果是（）A

a

5

B

a

6

C．

a

8

D．

a

9．不等式組

x，的解集是（）xA

x

BC．

x

D．

．用換元法解分式方程

時(shí)如果設(shè)xx

，將原方程化為關(guān)于y

的整式方程，那么這個(gè)整式方程是（）A

y

B

yC．

y

2

y

D．

y

2

y．拋物線

y2(x)

2

（m，是數(shù)）的頂點(diǎn)坐標(biāo)（）A

(，)

B

()

C．

(

D．

(．下列正多邊形中，中心角等于內(nèi)角的是（）A正六邊形正五邊形C．四形

C．三邊形AB∥CD∥EF，那么下列結(jié)論正確的是（）．如圖，已知BCDFABDFCEADC．DEFBE二填題本題12題，題4分滿48分

E

AC

圖

BD

F【將果線入題的應(yīng)置1．分母有理化_______5．方程

x

的根是．．如果關(guān)于

的方程

x

2

（

為常數(shù)）有兩個(gè)相等的實(shí)數(shù)根，那么

．知函數(shù)

f(x)

，那么f(3)．11反比例函數(shù)

圖像的兩支分別在第______象限．．拋物線

y

向上平移一個(gè)單位后，得以新的拋物線，那么新的拋物線的表達(dá)式是．．果從小明等6學(xué)生中任選名作為“世博會(huì)”志愿者，那么小明被選中的概率是．．商品的原價(jià)為100元如果經(jīng)過(guò)兩次降價(jià)，且每次降價(jià)的百分率都是，那么該商品現(xiàn)在的價(jià)格元（結(jié)果用含m的數(shù)式表示．圖2，在△中AD是的中線，設(shè)向量BC

A-第頁(yè)-/-共8頁(yè)-

B

D

C圖

如果用向量，b表向量，么AD．圓

中，弦

AB

的長(zhǎng)為6，它所對(duì)應(yīng)的弦心距為4，那么半徑

OA

．．四邊形

ABCD

中，對(duì)角線

AC

與

BD

互相平分，交點(diǎn)為

．在不添加任何輔助線的前提下，要使四邊形為矩形，還需添加一個(gè)條件，這個(gè)條件可以__________________在Rt△中AB，M為邊上點(diǎn)聯(lián)結(jié)AM（圖所果△沿線AM翻折后，點(diǎn)B恰

A落在邊

AC

的中點(diǎn)處，那么點(diǎn)

M

到

AC

的距離是．三解題本題題滿78分

B題滿分）

M計(jì)算：

aa

2

2a

．

圖

C題滿分）解方程組：2xxy.

①②題滿分，每小題滿分各5分如圖，在梯形ABCD中AD∥BCDC，結(jié)．（1）求tanACB的；（2）若M、N分是、的點(diǎn)，聯(lián)結(jié)MN，求線段MN的長(zhǎng)．題滿分分，第（1小題滿分2分第）小題滿分3分，第3）小題滿分2分第4小題滿分3分為了了解某校初中男生的身體素質(zhì)狀況，在該校六年級(jí)至九年級(jí)共四個(gè)年級(jí)的男生中，分別抽部分學(xué)生進(jìn)行“引體向上”測(cè)試．所有被測(cè)試者的“引體向上”次數(shù)情況如表一所示；各年級(jí)的被測(cè)試人占所有被測(cè)試人數(shù)的百分率如圖5所（其中六年級(jí)相關(guān)數(shù)據(jù)未標(biāo)出-第頁(yè)-/-共8頁(yè)-

次數(shù)人數(shù)

表一根據(jù)上述信息，回答下列問(wèn)題（直接寫(xiě)出結(jié)果（1）六年級(jí)的被測(cè)試人數(shù)占所有被測(cè)人數(shù)的百分率是；（2）在所有被測(cè)試者中，九年級(jí)的人是；（3有被測(cè)試者上不于的人數(shù)所占的百分率是；（4）在所有被測(cè)試者的“引體向上”數(shù)中，眾數(shù)是．

八年級(jí)25%七年級(jí)25%

九年級(jí)30%六年級(jí)圖題滿分，每小題滿分各6分已知線段ACBD相于點(diǎn)O，結(jié)、，為的點(diǎn)，F(xiàn)為OC的點(diǎn)，聯(lián)EF（圖示

A

D（1）添加條件

D

，

，

O求證：

ABDC

．（2）分別為①OFE為②DC”記為③，添加條件①、③，以②為結(jié)論構(gòu)成命題1，添加條件②、③，以①

B

E

圖

F

C為結(jié)論構(gòu)成命題2．命題1是命，命題2是命題（選擇“真”或“假”填入空格題滿分，每小題滿分各4分

y

在直角坐標(biāo)平面內(nèi)，

為原點(diǎn)，點(diǎn)

A

的坐標(biāo)為

，

DM

點(diǎn)

C

的坐標(biāo)為

(0

，直線

M∥x

軸（如圖7所

B

與

點(diǎn)

A

關(guān)于原點(diǎn)對(duì)稱，直線

yx（b常數(shù)）經(jīng)過(guò)點(diǎn)B，

B

O

A

x

與

直線-第頁(yè)-/-共8頁(yè)-圖

相交于點(diǎn)

D

，聯(lián)結(jié)

OD

．（1）求

b

的值和點(diǎn)

D

的坐標(biāo)；（2）設(shè)點(diǎn)

P

在

軸的正半軸上，若

△

是等腰三角形，求點(diǎn)

P

的坐標(biāo)；（3）在（）的條件下，如果以

PD

為半徑的圓

P

與圓

外切，求圓

的半徑．題滿分，第）小題滿分4分，第（2小題滿分分，第（）小題滿分）已知

ABC，BC∥，為線段上動(dòng)，點(diǎn)Q在射線AB上且滿足PCAB

（如圖示-第頁(yè)-/-共8頁(yè)-

△△（1）當(dāng)

AD

，且點(diǎn)

與點(diǎn)

B

重合時(shí)（如圖示線

的長(zhǎng)；3（2在中聯(lián)AP當(dāng)AD且在段AB上設(shè)點(diǎn)、之間的距離為，，2△PBC其中

APQ

表示

△APQ

的面積

△PBC

表示

△

的面積y

關(guān)于

的函數(shù)解析式寫(xiě)函數(shù)定義域；（3）當(dāng)AD，且點(diǎn)Q在線的長(zhǎng)線上時(shí)（如圖10所QPC的?。瓵

P

D

P

D

DPQB

圖

C

（Q

圖9

C

BQ

圖10

C九年級(jí)上數(shù)學(xué)摸底試卷答案說(shuō)明：．解答只列出試的一種或幾種解法．如果考生的解法與所列解法不同，可參照解中評(píng)分標(biāo)準(zhǔn)相應(yīng)評(píng)分；．第一、二大題無(wú)特別說(shuō)明，每題評(píng)分只有滿分或零分；．第三大題中各右端所注分?jǐn)?shù)，表示考生正確做對(duì)這一步應(yīng)得分?jǐn)?shù)；．評(píng)閱試卷，要持每題評(píng)閱到底，不能因考生解答中出現(xiàn)錯(cuò)誤而中斷對(duì)本題的評(píng)．如果考生的解答在-第頁(yè)-/-共8頁(yè)-

．;．；9；．；12某一步出現(xiàn)錯(cuò)誤，影響后繼部分未改變本題的內(nèi)容和難度，視影響的程度決定后繼部分的給分但原則上不超過(guò)后繼部分應(yīng)得分?jǐn)?shù)的半；．;．；9；．；12．評(píng)分時(shí)，給分扣分均以為基本單位．一選題本題題滿24分．；．；3．A;4B;．C;6A．二填題本題12題，分分１１４２．；．；．)；6

11一、三；1．b；2．；17BD（或

90

等

18.

三解題本題7題，滿78分）．：原式＝

2((aaa(

·········································7）＝

a

·······································································（）＝

1a

··············································································（）＝．················································································（1分．：由方程①得

y

，③························································）將③代入②，得

x

2

(x

，··········································（）整理，得

x

0

，·····························································（2分解得x，x·································································3分1分別將x，x入③，得y，··························（）1212所以，原方程組的解為···································（1）yy21．解）過(guò)A作AE，垂足為．···········································（）在Rt△中∵60AB，∴

BE60

，···············································（分ABBsin603

．················································（1分∵

12

，∴

8

．

·······························································（分在

Rt

△

中，

33EC8

．···································（）（2在梯形

ABCD

中，∵

ABDC

，

B60

，∴

DCB60

．·······································································（1分）過(guò)點(diǎn)

D

作

DF

，垂足為

F

，∵

DFCAEC

，∴

AE//DF

．∵AD，∴四邊形AEFD是行四邊形．∴AD．····················（）在Rt△DCF中FCDCFcos

，··················（）∴

．AD．∵

M

、

分別是

AB

、

的中點(diǎn)，∴

MN

ADBC22

．·······（2分））

20%

；··················································································

（2分（2；···················································································（）-第頁(yè)-/-共8頁(yè)-

（3

35%

；················································································（）（4

5

．······················································································（）23）證明

OEFOFE

，∴

OEOF

．···································································（1分∵為的點(diǎn)F為的中點(diǎn)，∴

，2OF

．············································（）∴

OBOC

．···································································（1分∵

AD

，

DOC

，∴△≌△DOC．······················································（）DC

．···································································（）（2真；·······················································································（3分假．···························································································（3分）24．解）∵的坐標(biāo)為，B點(diǎn)A關(guān)原對(duì)稱，∴點(diǎn)的標(biāo)為·······························································（）∵直線

y經(jīng)過(guò)點(diǎn)，，得·························（1分）∵點(diǎn)的標(biāo)為，直CM//軸∴設(shè)點(diǎn)的標(biāo)為(．·····（）∵直線

y與直線CM相于點(diǎn)D，∴的標(biāo)為．1分）（2∵D的坐標(biāo)為(3，∴OD．·············································（1分當(dāng)當(dāng)

PD時(shí)，點(diǎn)的標(biāo)為(6；···································（1分）POOD5時(shí)點(diǎn)的標(biāo)為(5，·····································（）當(dāng)PO時(shí)設(shè)點(diǎn)P的標(biāo)為x

，∴

x

(

2

2

，得

2525，∴點(diǎn)P的標(biāo)為(．··········（1分）66綜上所述，所求點(diǎn)

P

的坐標(biāo)是

(5或

256

．（3當(dāng)以

PD

為半徑的圓

P

與圓

外切時(shí)，若點(diǎn)

P

的坐標(biāo)為

，圓

的半徑

PD

，圓心距

PO

，∴圓的徑r．····································································（）若點(diǎn)

P

的坐標(biāo)為

，圓P的徑PD

，圓心距

PO5

，∴圓

的半徑

r

．························································（2分）綜上所述，所求圓的徑等于1或525．解）∵BC，∴ADBDBC

．．∵

AD2

，∴

ADB

．∴

DBC

．∵

ABC

．∴

PBC

．················································（1分∵

，AB，Q與點(diǎn)重合，∴PCAB

PBPC

．∴

PCBPBC45

．·····························································（）∴

BPC

．·········································································（）在Rt△中，PCBCC45

32

．····················（1分）（2過(guò)點(diǎn)

P

作

PE

，

PFAB

，垂足分別為

E

、

F

．···················（）∴

FBEBEP

．∴四邊形是矩形．∴//，PEBF．-第頁(yè)-/-共8頁(yè)-

△PBCAP△PBCAP∵

AD//BC

，∴