山西省陵川第一中學、高平一中、陽城一中2020學年高二數(shù)學上學期第三次月考試題 理(含解析)_第1頁
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山西省陵川第一中學、高平一中、陽城一中2020學年高二上學期第三次月考數(shù)學(理)試題一、選擇題(本大題共12小題,共60.0分)1.設集合,則集合中元素的個數(shù)是( )A. B. C. D. 【答案】C【解析】A=0,1,2,B=xy|xA,yA,當x=0,y分別取0,1,2時,xy的值分別為0,1,2;當x=1,y分別取0,1,2時,xy的值分別為1,0,1;當x=2,y分別取0,1,2時,xy的值分別為2,1,0;B=2,1,0,1,2,集合B=xy|xA,yA中元素的個數(shù)是5個故選C2.已知命題,其中正確的是( )A. 使B. 使C. 使D. 使【答案】D【解析】【分析】由特稱命題的否定為全稱命題即可得解【詳解】命題,為特稱命題,其否定為全稱命題,所以使.故選D.【點睛】本題主要考查了含有量詞的命題的否定,由全稱命題的否定為特稱命題,特稱命題的否定為全稱命題即可得解.3.已知方程表示橢圓,則的取值范圍為A. B. C. D. 【答案】D【解析】【分析】令兩分母均大于零且不相等解出m的范圍即可【詳解】方程表示橢圓,解得且故選:D【點睛】本題考查了橢圓的方程,熟記方程的特征,準確計算是關鍵,屬于基本知識的考查4.直線的傾斜角的變化范圍是A. B. C. D. 【答案】D【解析】【分析】由已知直線方程求出直線斜率的范圍,再由斜率與直線傾斜角的關系得答案【詳解】由,得此直線的斜率k=設其傾斜角為,則故選:D【點睛】本題考查直線的傾斜角,考查了直線的傾斜角與斜率的關系,熟記正切函數(shù)性質(zhì)是關鍵,是基礎題5.已知的平面直觀圖是邊長為1的正三角形,那么原的面積為A. B. C. D. 【答案】A【解析】【分析】由直觀圖和原圖像的面積比為易可得解.【詳解】直觀圖ABC是邊長為1的正三角形,故面積為,而原圖和直觀圖面積之間的關系,那么原ABC的面積為: ,故選A.【點睛】本題主要考查平面圖形的直觀圖和原圖的轉(zhuǎn)化原則的應用,要求熟練掌握斜二測畫法的邊長關系,比較基礎直觀圖和原圖像的面積比為掌握兩個圖像的變換原則,原圖像轉(zhuǎn)直觀圖時,平行于x軸或者和軸重合的長度不變。平行于y軸或者和軸重合的線段減半。原圖轉(zhuǎn)直觀圖時正好反過來,即可.6.若圓與圓關于直線對稱,則直線的方程是A. B. C. D. 【答案】D【解析】【分析】由題意化圓C為標準方程,由兩圓位置關系得兩圓相交,直線l是兩圓的公共弦所在的直線,故把兩圓的方程相減可得直線l的方程【詳解】由題圓C:,則兩圓心距為,故兩圓相交由于圓O:與圓C:關于直線l對稱,則直線l是兩圓的公共弦所在的直線,故把兩圓的方程相減可得直線l的方程為,故選:D【點睛】本題主要考查圓和圓的位置關系,直線與圓的位置關系的應用,判斷直線l是兩圓的公共弦所在的直線,是解題的關鍵,屬于中檔題7.在平面直角坐標系中,雙曲線中心在原點,焦點在軸上,一條漸近線方程為,則它的離心率為A. B. C. D. 【答案】A【解析】試題分析:因為雙曲線中心在原點,焦點在軸上,一條漸近線方程為,所以=2,=。故選A??键c:本題主要考查雙曲線的幾何性質(zhì)。點評:易錯題,在雙曲線問題中,涉及a,b,c,e關系的考題經(jīng)常出現(xiàn),本題中要分清焦點所在坐標軸,以準確求離心率。8.已知直線和平面,且在,內(nèi)的射影分別為直線和,則和的位置關系是A. 相交或平行B. 相交或異面C. 平行或異面D. 相交、平行或異面【答案】D【解析】解:因為直線和平面,且在內(nèi)的射影分別為直線和,則和的位置關系可能有3種情況。9.如圖,三棱錐的底面為正三角形,側(cè)面與底面垂直且,已知其正視圖的面積為,則其側(cè)視圖的面積為( )Failed to download image : /QBM/2020/3/23/1572554115211264/1572554120732672/STEM/68c315e7099040ebb682edfbdad85a83.pngA. B. C. D. 【答案】B【解析】試題分析:設底面正ABC的邊長為a,側(cè)面VAC的底邊AC上的高為h,則底面正ABC的高為,平面VAC為正視圖的投影面,;左視圖的高與主視圖的高相等,左視圖的高是h,又左視圖的寬是底面ABC的邊AC上的高,考點:簡單空間圖形的三視圖10.橢圓的焦點為和,點在橢圓上,如果線段的中點在軸上,那么是的A. 7倍B. 5倍C. 4倍D. 3倍【答案】A【解析】本題考查橢圓定義,幾何性質(zhì),平面幾何知識及運算.因為線段的中點在軸上,是的中點,所以的邊 即時直角三角形,且Failed to download image : /QBM/2020/8/24/1570289107230720/1570289112154112/EXPLANATION/85613fe585be40679b59bd3b2a435d09.png由橢圓定義得:又由(1),(2)解得故選A11.已知高為3的正三棱柱的每個頂點都在球的表面上,若球的表面積為,則異面直線與所成角的余弦值為A. B. C. D. 【答案】B【解析】【分析】由三棱柱外接球的表面積得:三棱柱的底面邊長為a,則此三棱柱的外接球的半徑,又由,所以,得:,由異面直線平面角的作法得:分別取BC、的中點E、F、G,連接GF、EF、EG,因為,則或其補角為異面直線與所成角,再利用余弦定理求解即可【詳解】Failed to download image : /QBM/2020/4/4/2175117180534784/2180641781104640/EXPLANATION/6e19d7708ef34e358cd5eef7e24617d1.png設三棱柱的底面邊長為a,則此三棱柱的外接球的半徑,又由已知有,所以,聯(lián)立得:,分別取BC、的中點E、F、G,連接GF、EF、EG,因為,則或其補角為異面直線與所成角,又易得:,在中,由余弦定理得:,又為銳角即異面直線與所成角的余弦值為,故選:B【點睛】本題考查了三棱柱外接球的表面積及異面直線平面角的作法,熟記正三棱柱外接球性質(zhì),準確作出異面直線所成角是關鍵,屬中檔題12.某幾何體中的一條線段長為,在該幾何體的正視圖中,這條線段的投影是長為的線段,在該幾何體的側(cè)視圖與俯視圖中,這條棱的投影分別是長為和的線段,則的最大值為A. B. C. 4D. 【答案】C【解析】試題分析:結(jié)合長方體的對角線在三個面的投影來理解計算,如圖設長方體的高寬高分別為,由題意得:, ;,所以 , ,當且僅當時取等號.故選C.Failed to download image : /QBM/2020/11/8/1571387532812288/1571387538333696/EXPLANATION/27426edb51d440c7a2e628d5094a1061.png考點:1.三視圖;2.均值不等式二、填空題(本大題共4小題,共20.0分)13.已知兩條平行直線,間的距離為2,則_【答案】38或-2【解析】將l1:3x+4y+5=0改寫為6x+8y+10=0,因為兩條直線平行,所以b=8由 =2,解得c=30,或c=-10,所以38或-2.故答案為38或-214.若,則_【答案】64【解析】【分析】根據(jù)換底公式即可根據(jù)條件得出:,從而可得出,解出x即可【詳解】;故答案為:64【點睛】考查對數(shù)的運算性質(zhì),以及對數(shù)的換底公式,對數(shù)的定義,指數(shù)運算,熟記運算性質(zhì)是關鍵,是基礎題15.如圖,二面角的大小是60,線段.,與所成的角為30.則與平面所成的角的正弦值是 .Failed to download image : /QBM/2020/6/18/1569763478052864/1569763482959872/STEM/3a287d099a7c45889daab1bbcda9b4c9.png【答案】【解析】試題分析:過點A作平面的垂線,垂足為C,在內(nèi)過C作l的垂線垂足為D,連接AD,有三垂線定理可知ADl,故ADC為二面角-l-的平面角,為60,又由已知,ABD=30,連接CB,則ABC為AB與平面所成的角設AD=2,則AC=,CD=1AB=4sinABC=;故答案為。Failed to download image : /QBM/2020/3/11/1571133611581440/1571133616988160/EXPLANATION/8f1461da95b94e4faa37ef88541fffb9.png考點:本題主要考查二面角的計算。點評:基礎題,本解法反映了求二面角方法的“幾何法”“一作、二證、三計算”。16.設橢圓的右頂點為、右焦點為為橢圓在第二象限上的點,直線交橢圓于點,若直線平分線段,則橢圓的離心率是_【答案】【解析】試題分析:如圖,設AC中點為M,連接OM,則OM為的中位線,于是 ,且,即Failed to download image : /QBM/2020/11/6/1572281591824384/1572281597788160/EXPLANATION/1293316472f84015b6f6a3e452476181.png考點:橢圓的離心率三、解答題(本大題共6小題,共70.0分)17.已知關于的方程有兩個不等的負根;關于的方程無實根,若為真,為假,求的取值范圍.【答案】解:若方程有兩個不等的負根,則,解得,即P:Failed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/fb6e8d814ed644b6b64fa74bd5ca0033.pngFailed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/fb6e8d814ed644b6b64fa74bd5ca0033.pngFailed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/fb6e8d814ed644b6b64fa74bd5ca0033.pngFailed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/fb6e8d814ed644b6b64fa74bd5ca0033.png3分若方程Failed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/eec70d619e3f47f8a6a4dd89b8720c33.png無實根,則Failed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/297a7b079dc64b6aa6e97d5ce9706c09.png,解得,即q:Failed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/fb6e8d814ed644b6b64fa74bd5ca0033.pngFailed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/fb6e8d814ed644b6b64fa74bd5ca0033.pngFailed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/fb6e8d814ed644b6b64fa74bd5ca0033.pngFailed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/fb6e8d814ed644b6b64fa74bd5ca0033.png6分因Failed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/7c9273d5d952449483f53080df4905fb.png為真,F(xiàn)ailed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/1ac9db80a46e4209b8dcc423e5f1af5b.png為假,所以p、q兩命題中應一真一假,即p為真,q為假或q為真,p為假Failed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/fb6e8d814ed644b6b64fa74bd5ca0033.pngFailed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/fb6e8d814ed644b6b64fa74bd5ca0033.pngFailed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/fb6e8d814ed644b6b64fa74bd5ca0033.pngFailed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/fb6e8d814ed644b6b64fa74bd5ca0033.png8分Failed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/3139d4e11aed4f43b86efcc67d7e251d.png或Failed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/72b4267164ef4c46a3ca9fc533bb5a32.png,解得Failed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/4a4e15be8ee74305ae6f394b2c3825e6.png或Failed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/0ba0d9441fef4cc5a557b77e454b1f98.pngFailed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/fb6e8d814ed644b6b64fa74bd5ca0033.pngFailed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/fb6e8d814ed644b6b64fa74bd5ca0033.pngFailed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/fb6e8d814ed644b6b64fa74bd5ca0033.pngFailed to download image : /QBM/2020/1/16/1570692550909952/1570692556283904/ANSWER/fb6e8d814ed644b6b64fa74bd5ca0033.png12分 【解析】試題分析:由題意可得命題p為真時,m2,命題q為真時,1m3, 由于pq為真命題,pq為假命題,則p,q一真一假,據(jù)此分類討論可得實數(shù)m的取值范圍是(1,23,+)試題解析:命題p為真時,實數(shù)m滿足=m240且m0,解得m2, 命題q為真時,實數(shù)m滿足=16(m2)2160,解得1m3, 由于pq為真命題,pq為假命題,p,q一真一假;若q真且p假,則實數(shù)m滿足1m3且m2,解得1m2; 若q假且p真,則實數(shù)m滿足m1或m3且m2,解得m3 綜上可知實數(shù)m的取值范圍是(1,23,+)18.已知正方體,是底對角線的交點求證:Failed to download image : /QBM/2020/4/4/2175117180534784/2180641781653504/STEM/4ab2c1b4f3e64d0c839ea36d609db80b.png面;面【答案】見解析?!窘馕觥吭囶}分析:(1)取 的邊的中線 ,由證四邊形 是平行四邊形,得,由線面平行的判定定理可得結(jié)論;(2)由 證得面 ,可得面 面 。(I)連結(jié)Failed to download image : /QBM/2020/3/7/1638718897512448/1638915824066560/EXPLANATION/8cf009aac7824db28fa691b4f03d2633.png,設Failed to download image : /QBM/2020/3/7/1638718897512448/1638915824066560/EXPLANATION/e872ddb37de042e3878c6944a88f5227.png 連結(jié)Failed to download image : /QBM/2020/3/7/1638718897512448/1638915824066560/EXPLANATION/97d6d190875840c2bc95f802feb6d401.png,F(xiàn)ailed to download image : /QBM/2020/3/7/1638718897512448/1638915824066560/EXPLANATION/0f4e4883a6774877958cfe8890208e93.png是正方體 四邊形是平行四邊形 Failed to download image : /QBM/2020/3/7/1638718897512448/1638915824066560/EXPLANATION/33f02047520b42838c417452f4cd468f.pngA1C1AC且Failed to download image : /QBM/2020/3/7/1638718897512448/1638915824066560/EXPLANATION/1632871fee7644afb2b6254711fd2426.png 又Failed to download image : /QBM/2020/3/7/1638718897512448/1638915824066560/EXPLANATION/8f038a512ec945afb536166e9c5cd4a3.png分別是Failed to download image : /QBM/2020/3/7/1638718897512448/1638915824066560/EXPLANATION/37c856e3095241839eb822cf93e3444a.png的中點,且Failed to download image : /QBM/2020/3/7/1638718897512448/1638915824066560/EXPLANATION/ce03b27159514c09a43587a0841bbef5.png,四邊形是平行四邊形 ,面Failed to download image : /QBM/2020/3/7/1638718897512448/1638915824066560/EXPLANATION/ed855b22d56c4cf3bf4a5784ad841f60.png,F(xiàn)ailed to download image : /QBM/2020/3/7/1638718897512448/1638915824066560/EXPLANATION/7295a6a0e67c4c8988a0c0b6847c5c63.png面,面Failed to download image : /QBM/2020/3/7/1638718897512448/1638915824066560/EXPLANATION/ed855b22d56c4cf3bf4a5784ad841f60.png (II)在正方體中,AA1平面A1B1C1D1,平面A1B1C1D1, 在平面A1B1C1D1內(nèi),,,, ,面A1C面AB1D1 點睛:處理直線、平面平行問題時應注意的事項(1)在推證線面平行時,一定要強調(diào)直線不在平面內(nèi),否則,會出現(xiàn)錯誤。(2)把線面平行轉(zhuǎn)化為線線平行時,必須說清經(jīng)過已知直線的平面與已知平面相交,則直線與交線平行。(3)兩個平面平行,兩個平面內(nèi)的所有直線并不一定相互平行,它們可能是平行直線、異面直線。19.為數(shù)列的前項和,已知,求數(shù)列的通項公式;設,求數(shù)列的前項的和【答案】(1);(2).【解析】【分析】由題知,為數(shù)列的前n項和,可得兩式相減可得:,利用等差數(shù)列的通項公式即可得出由得,則,利用裂項求和方法即可得出【詳解】由題知,為數(shù)列的前n項和,則兩式相減可得,整理可得,又,則,則有,當時,解得或舍去則數(shù)列是以3為首項,2為公差的等差數(shù)列,則綜上所述,數(shù)列的通項公式為;由得,則則數(shù)列的前n項的和為: 綜上所述,數(shù)列的前n項的和為【點睛】本題考查了數(shù)列遞推關系、等差數(shù)列的通項公式與求和公式、裂項求和方法,考查了推理能力與計算能力,屬于中檔題20.在中,角所對的邊分別為,求角的大?。蝗?,的面積為,求及的值【答案】(1);(2).【解析】【分析】利用二倍角公式,化簡可求得的值,進而求C利用余弦定理可求得c與a的關系,進而求得,然后利用三角形面積公式和已知等式求得c【詳解】,即,.【點睛】本題主要考查了正弦定理和余弦定理的應用,二倍角公式,面積公式,在解三角形的問題中應靈活運用余弦和正弦定理實現(xiàn)邊角的轉(zhuǎn)化,是中檔題21.如圖1,在等腰直角三角形中,分別是上的點,為的中點將沿折起,得到如圖2所示的四棱椎,其中證明:平面;求二面角的平面角的余弦值Failed to download image : /QBM/2020/4/4/2175117180534784/2180641782767616/STEM/37d7218a5d36428ca47d48957f60a788.png【答案】(1)詳見解析 (2)【解析】試題分析:(1)F為ED的中點,連接OF,AF,根據(jù)已知計算出Failed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/31f90e25a8114aabba606d5b07c8e5e9.png的長度,滿足勾股定理,F(xiàn)ailed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/98471d41d7944facbad8758e8f3202e4.png, AF為等腰ADE底邊的中線,F(xiàn)ailed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/41fec88d699344148a11dbba7b767e19.png,Failed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/74599d1dff9545b1bd9ac2689fd3a60c.png,證得線面垂直,線線垂直,再線面垂直;(2)過點O作Failed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/64859c08d76d4b62a2b2da3b6beee7cc.png的延長線于Failed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/dcdc6b14e9b342d1925fedb21cd1911c.png,連接Failed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/e0e1aa7d6e14441288fc81d6846eb48e.png利用(1)可知:Failed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/f47662e4ae984b59b061d73be381ff28.png平面Failed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/bd80f1286f454971a9560116b2b27383.png,根據(jù)三垂線定理得Failed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/9a75faa99d384330a724fe6aadbc709b.png,所以Failed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/b2eb1dec827d4b6281af11202137192e.png為二面角Failed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/6e1c501aadfb45768176612773322aba.png的平面角在直角Failed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/17f2389369754e43b45dfbe70a9c5612.png中,求出Failed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/838c06fca6ee424f84912f93227fa840.png即可;試題解析:證明: (1)設F為ED的中點,連接OF,AF,計算得AF=2,OF=1Failed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/b7bebe78d2a74489a97172fd449bb057.pngAF為等腰ADE底邊的中線,AFDEOF在原等腰ABC底邊BC的高線上,OFDE又AF,OF平面AOF, AFOF=F,DE平面AOFAO平面AOF, DEAO在AFO中,A+Failed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/07d37c4170224985ae8775e0dd69f13c.png=3+1=Failed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/72e48b392940468bbe3c819493f5c105.png,AOOFOFDE=F,OF平面BCDE,DE平面BCDE, AO平面BCDE 6分(2):如答圖1,過O作CD的垂線交CD的延長線于M,連接AMAO平面BCDE,CD平面BCDE, CDAO OMAO=O, CD平面AOMAM平面AOMCDAM Failed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/75b44cf472dc4ed0a655c656d2548bd5.pngAMO為所求二面角的平面角在RtOMC中,OM=Failed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/84ca5407d57b4d0985112603e1f7ae33.png=Failed to download image : /QBM/2020/4/23/1571639446421504/1571639452049408/EXPLANATION/2ce77d4a7baa42f99f1185703d1f86

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