蘇科版八年級(jí)數(shù)學(xué)下冊(cè)第9章-第11章綜合測(cè)試

22一選題本大題共小題，每小題3分，共18分在小題所給出的四個(gè)選項(xiàng)中，只有一項(xiàng)是符合題目要求的，請(qǐng)將確選項(xiàng)的字母代號(hào)填涂在答題卡相應(yīng)位置上）1．下圖形是中心對(duì)稱圖形的是·············································································（）ABCDx2．若式有意義，則x滿的條件是···························································（）xA．x0BxC．x3．≥．下列函數(shù)中，是反比例函數(shù)的為（）A．y2x

B．y

2x

C．

y

3x

D．2y23．在代數(shù)式，，，中分式有（）2aA2個(gè)B3C個(gè)D5個(gè)5．下等式成立的是·································································································（）231abaA．+＝；．＝；．＝；．＝-ababa+ba+bab-b-b-a+ba+b63．某反比例函數(shù)的圖象經(jīng)過點(diǎn)，此數(shù)的圖象也經(jīng)過點(diǎn)（）A，-3）B-3，）C，3）D，6菱具有而一般平行四邊形不具有的性質(zhì)是

()A對(duì)邊相等對(duì)角相等C．對(duì)角線相平分；D．對(duì)角線互相垂如，在平行四邊形中下結(jié)論中錯(cuò)誤的是（）A∠1=∠B．∠=∠BCDC=CDD．AC⊥BD下四個(gè)命題：①一組對(duì)邊平行且一組對(duì)角相等的四邊形是平行四邊形；②對(duì)角線互相垂直且相等的四邊形是正方形順次連結(jié)矩形四邊中點(diǎn)得到的四邊形是菱形等三角形既是軸對(duì)稱圖形又是中心對(duì)稱圖形．其中真命題共有（）A．1個(gè)B2個(gè)C．3個(gè)D．個(gè)10．圖，在四邊形中∠A90°＝，＝7，點(diǎn)、分為線段BC、上的動(dòng)點(diǎn)，點(diǎn)、分為DM、的中點(diǎn)，則長(zhǎng)度的最大值為···················（）A．

B．．D2二填題本大題共8小題，每小題3分共24分.不需寫解答過程，請(qǐng)將答案直接寫在答題卡相應(yīng)位置上）x11．若式的為0則x＝▲．1-．分式、的簡(jiǎn)公分母是▲．3x13．形中，對(duì)角線＝，＝，菱形的面積是▲．14如圖，矩形ABCD的角線AC、交于點(diǎn)O，∠AOD，＝4，則長(zhǎng)▲．ADA

E

O

D

A

P

D

A

DB

O

FBC

B

B

P

E

F第14題圖

第15題圖

第17題圖

第題圖15圖eq\o\ac(□,)的角線相于點(diǎn)點(diǎn)F別是線段的點(diǎn)＋＝22，的周長(zhǎng)是16cm，則的長(zhǎng)為▲cm．16．知x0，分式的是▲．x圖形的長(zhǎng)為別是邊的上點(diǎn)MC＝MB＝NC是對(duì)角上上點(diǎn)，則PM的最小值是▲．．如圖所示，點(diǎn)為正方的對(duì)角線上一點(diǎn)，過點(diǎn)P作BC，垂足分別為點(diǎn)E、，連接EF．下列結(jié)論中是腰直角三角形；AP；AD＝；PFE＝BAP其中正確的結(jié)論是▲）三解題本題共小題，共76分，請(qǐng)?jiān)诖鹂ㄖ付▍^(qū)域內(nèi)作答，解答時(shí)應(yīng)寫出文字說明、推理過程或演算步驟）19）算．（1-+

+b

；

（2

a．a(chǎn)a20

）圖，點(diǎn)A、是的角線EF所在直線上的兩點(diǎn)且AE＝CF．求證：四邊形平行四邊形．D

FEAB第20題圖xx1216）先化簡(jiǎn)：)，后在－，，2四數(shù)中找xx一個(gè)你認(rèn)為合適的代求值．226）方程：

x

31

．236觀等11＝-，……45

111111＝1-②＝-＝-④′22′3344′5試用含字母的式表示出你發(fā)現(xiàn)的規(guī)律，并證明該等成立；1111+++＝）227本10圖27本10圖1圖248）圖，一次函數(shù)y=+b與比例函數(shù)=的圖象交于點(diǎn)A（16B（3n）兩點(diǎn)．求反比例函數(shù)和一次函數(shù)的表達(dá)式；點(diǎn)是一次函數(shù)=kx+圖位第一象限內(nèi)的一點(diǎn)，過點(diǎn)作MNx軸，垂足為點(diǎn)N過點(diǎn)作BDy軸，垂足點(diǎn)D，eq\o\ac(△,)MON的積小eq\o\ac(△,)BOD的面積，直接寫出點(diǎn)M的坐標(biāo)x取值范圍．第題258）eq\o\ac(□,)中過點(diǎn)D作DE⊥于，點(diǎn)在CD上，＝BE，連接，．求證：四邊形DEBF是矩形；若平分∠，＝，＝，求的面積．

A

DFBE第題圖2610）圖1，方ABCD中點(diǎn)O是對(duì)角ACAO上（不與A、重）的一個(gè)動(dòng)點(diǎn)，過點(diǎn)P作⊥PB且交于點(diǎn)．求證：＝；過點(diǎn)E作于F，圖2．正方形ABCD的長(zhǎng)為2，在點(diǎn)P運(yùn)的過程，的度是否發(fā)生變化？若不變，請(qǐng)直接寫出這個(gè)不變的值；若變化，請(qǐng)說明理由．ADAPP

DB

O

EOF

E（把一張矩形片ABCD如圖方式折疊使點(diǎn)落邊AD（為點(diǎn)）點(diǎn)落在點(diǎn)分與邊、BC交于試在圖中連接，求證：四邊形是形；若AB，＝，求線能取到的整數(shù)值．A

B

ADA

B

F

C

B

C

B

C第27題圖

備用圖

備用圖121212122810）面直角坐標(biāo)系xOy中已知函數(shù)=（＞0）與y﹣（＜）的圖象如圖所示，點(diǎn)AB是數(shù)=（x＞）圖象上的兩點(diǎn)點(diǎn)P是=（x＜）的圖象上的一點(diǎn)且∥x軸點(diǎn)Q是x軸一點(diǎn)設(shè)點(diǎn)B的坐標(biāo)分別為mm求的面積；若是等腰直角三角形，求點(diǎn)Q的標(biāo)；若是以AB為的等腰三角形，求的．參答及分準(zhǔn)一選擇（330A

C

C

A

C

A

D

D

B

D二填題22011．；．

y

；；．；．2.5；．17．；．②④．三解題19）原＝

(a-ba)+a+b

··························································································2＝

a+ba+b

．·········································································································3（）式＝

a(a-×(+1)a-

··································································································2＝

aa+

．·············································································································320．明：連接DB交EF于O．∵四邊形是行四邊形，∴＝OB，OE＝．·······································································································2∵＝，∴＋＝＋，＝．4∴四邊形是平行四邊形．·······················································································21)(x-1)x-21．：原式＝···············································································x-+x＝+．·············································································································4取x＝．····························································································································5∴原式2＋＝．············································································································6（222．；驗(yàn)不能漏。123）＝-（為正整·······················································nn+1)eq\o\ac(□,S)eq\o\ac(□,S)證明：∵

1+1n-＝-nn+(+1)

n+-n＝＝．······················4n(+n(n+∴

111＝-．n(n++（2

．···························································································6（124）

y

6x

，y（）x3

；25）四邊形是行四邊形，∥AB，即DF∥EB．又∵DF＝，四邊形DEBF是行四邊形．2∵⊥，∴∠＝．四邊形DEBF是形．（）四邊形DEBF是矩形，∴＝＝，BDDF．

4∵⊥，＝

+DE

＝

+

＝．∵∥，∠＝．∵平∠DAB，∠DAF＝∠∴∠＝∠．∴＝＝．···········································································································∴＝．∴ABAEBE＝＋＝．∴＝·BF＝×＝．·············································································826）圖，連接．∵四邊形是方形，∴，∠＝DCA，∠BCD＝．又∵PC＝，△BCP≌DCP．∴＝，PBCPDC．3∵⊥，BPE＝．∴在四邊形BCEP中∠∠＝－∠－∠＝．又∵∠＋∠，∴∠＝PED∴∠＝∠．∴＝．·················································································································6∴＝．·················································································································7（2P3BCCD11（）

PE

的長(zhǎng)度不發(fā)生變化，＝．······································································10（OBPEFeq\o\ac(△,≌)82）連接BB′．由折疊知點(diǎn)B、關(guān)EF稱．是段BB的垂直平分線．＝BEBFF．································································································2四形是矩形ADBC．B＝．由折疊得B＝．B＝B．B′＝BF················································································································＝BE′F＝．四形BFB是菱形．

4（）圖1，點(diǎn)E與重時(shí)，四邊形′是正方形，此時(shí)BF最．5∵四邊形ABFB′正方形，∴＝＝，最為．··············································································6如圖2，點(diǎn)與D重時(shí)BF最大．·····························································7設(shè)BF，＝-x，＝＝．在eq\o\ac(△,Rt)中，由勾股定理得＋2＝．∴-x)+＝x2，得，即＝10．···················································9∴≤≤．∴線段長(zhǎng)能取到的整數(shù)值為8，，．·························································12eq\o\ac(△,S)1212eq\o\ac(△