y随着x的增大而增大.doc

y随着x的增大而增大?????11110.(1)当P=时,y=x+,即y=.100,xx,50,,2221?y随着x的增大而增大,即P=时,满足条件(?)……3分21又当x=20时,y==~100之间,所以新数据都在60~100之间,即,,1005021满足条件(?),综上可知,当P=时,这种变换满足要求;……6分2(2)本题是开放性问题,:(a)h?20;(b)若x=20,100时,y的对应值m,n能落在60~100之间,=20,y=,……8分axk,,20,,?a>0,?当20?x?100时,y随着x的增大…10分令x=20,y=60,得k=60?2令x=100,y=100,得a×80+k=100?1,a,12,由??解得,?.………14分yx,,,2060,,160,160,k,60,:(1)相等理由是:因为四边形ABCD、EFGH是矩形,所以SSSSSS,,,,,,,,,,,ECPCGQCGM所以,即:…………SSSSSS,,,,,,SS,,,,,,,CGQ34(2)AB=3,BC=4,AC=5,设AE=x,则EC=5-x,PCxMCx,,,(5),,551212122所以,即SPCMCxx,,,(5)Sxxx,,,,,(05)2525512552配方得:Sx,,,,()3,所以当时,x,2252S有最大值35(3)当AE=AB=3或AE=BE=或AE=,是等腰三角形.,:(1)在中,,?ABCACBC,.?,,,,,,BAACB36108,在与中,;,,,,AB36?ABC?CAD2,ACABAD,ACABAB.?,,ADACBC1????AD.?,,,,,,CDBDCB721083672,??ADC?BDC3636727222(2)设,则,,,,11xx,,,10ACx,,,1083636,,,1551解得3636xx,?,,(负根舍去).22(有8个等腰三角形),:(1)抛物线的对称轴x,,,22ay(2)A(30),,B(54),C(04),BCM12把点Ayaxax,,,54坐标代入中,解得Aa,,N160xQ1P153K2?,,,,yxx466P(3),,作轴于,易得,,,BQx,BQ,4AQ,8QBAN,?以?PAB为腰且顶角为角的有1个:.ABA?PAB122222?,,,,,ABAQBQ848019922222在Rt?ANPPNAPANABAN,,,,,,,80()中,1112,,5199?,P,,,1,,22,,?以?PAB为腰且顶角为角的有1个:.ABB?PAB2252952222在MPBPBMABBM,,,,,,,80Rt?BMP中,22242,,58295,?P,,,2,,22,,?以?PAB为底,顶角为角的有1个,?PAB3画P的垂直平分线交抛物线对称轴于,?3过点RtRt???PCKBAQPPK作垂直轴,垂足为,?,,.CKAQ2PK,?,CK5OK,13?,P(),:(1).PQQE,(2)?;?.(03),(66),?画图,:方法一:,?PEAEAP,,,65,Rt?APEFy1.?PFPE,,35218Q3,,?,,,,QPFEPA90?D,,,,AEPEPA90?3C12.?,,,QPFAEP3Q2G又,?,,,,EAPQFP90?36QP.????QPFPEA1E3MQPPFx61218243?,.PEEAB0(A)PEPF?.?QP,,153EA.?,Q(1215)3方法二:过点作,垂足为,,EGAPGE3,.?GP,6EG,12设,,QEQPx,,,6333222在?EQEGQG,,中,.Rt?QEG333222?(6)12xx,,,..?x,9.?QP,1253?,Q(1215).3(3):yxx,,3(026).??:(1)?,,,,AOC60AOAC,?是等边三角形.?AOC?.,,OAC60(2)?CP与相切,A?.,,ACP90?.,,,,,APCOAC90303又?(4,0),?.?.AACAO,,4PAAC,,28?.POPAOA,,,,,844(3)?过点作,垂足为,延长交于,CPOB,PCPQAC1111?OCOQ,是半径,?,?,OCOQ,OA11?是