2022年数学同步练习题考试题试卷教案中考数学专题讲座方程观点解几何计算题 .pdf

2009 中考数学专题讲座方程观点解几何计算题概述：含有未知数的等式便是方程，代数方面的应用题，?几何方面的计算题便是求某些未知数的值，都可用方程的观点去解决，一般一个未知数列一个方程，?两个未知数列两个方程典型例题精析例 1有一块直角三角形纸片，两直角边AC=6cm，BC=8cm，现将直角边AC?沿直线 AD折叠，使它落在斜边AB上，且与AE重合，求 CD长分析：RtABC，C=90，AC=6，BC=8 AB=10 由题意知ACD AEDDEB=90，DECD，AC=AE=6，设 CD=x，则 DE=x，而 EB=4，一个未知数，需要一个方程，从何而来，图中有直角，用勾股定理，有等式，有方程在 Rt DEB中，（8-x）2=x2+42，64-16x+x2=x2+16，16x=48，x=3（cm）例 2已知 O中，两弦AB、CD相交于 E，若 E为 AB中点，且CE：ED=1：4，AB=4，求 CD长解：CE：ED=1：4，设 CE=x，则 ED=4x，由相交弦定理得 CEED=AE EB，即 x4x=2 2，4x2=4，x=1CD=x+4x=5x=5例 3如图，AB为 O的直径，P点在 AB延长线上，PM切 O于 M点，若 OA=a，PM=3a，求 PMB的周长分析：条件符合切割线定理，设BP=x，则由PM2=PB PA（方程出来了）得（3a）2=x（x+2a），x2+2ax-3a2=0，OBCAED8-x6xBCAxED4aaMOBAP（x+3a）（x-a）=0，x1=a，x2=-3a（舍去）x=a，即 BP=a，连结 MO（常作辅助线）则 OMP=90，OB=BP=a，则 MB为 RtOMP 的斜边上的中线，MB=12OP=a MBP 的周长为2a+3a例 4如图，圆心在Rt ABC斜边 AB上的半圆切直角边AC、BC于 M、N，?其中 AC=?6，BC=8，求半圆的半径分析：设半径为R，（一个未知数建立一个方程即可），连 OM、ON、OC，则 OM=ON=R，用面积，SAOC+SBOC=SABC，得 6R+8R=6 8（一元一次方程）14R=48，R=247中考样题训练:1如图，在 ABC中，C=90，BAC=30，BC=1，D为 BC边上的一点，tan ADC是方程 3（x2+21x）-5（x+1x）=2 的一个根，求CD的长2如图，已知直线BC切 O于 C，PD为 O的直径，BP的延长线与CD?的延长线交于点A，A=28,B=26，求 PDC的度数MOBCANBCADOBCADP文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 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ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P23已知，如图，C为半圆上一点，ACCE，过 C 作直径的垂线CP，P为垂足，弦AE分别交 PC，CB于点 D，F（1）求证：AD=CD；（2）若 DF=54，tan ECB=34，求 PB的长EOBCADPF 4已知关于x 的方程 x2-（k+1）x+14k2+1=0 的两根是一个矩形两邻边的长（1）k 取何值时，方程有两个实数根；（2）当矩形的对角线长为5时，求 k 的值文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 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ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2 5如图所示，AB是 O的直径，BC是 O的弦，O?的割线 PDE?垂直 AB于点 F，交 BC于点 G，连结 PC，BAC=BCP，求解下列问题：（1）求证：CP是 O的切线；（2）当 ABC=30，BG=23，CG=43时，求以 PD、PE的长为两根的一元二次方程（3）若（1）的条件不变，当点 C在劣弧 AD上运动时，应再具备什么条件可使结论BG2=BF DO成立？试写出你的猜想，并说明理由EOBCADPGF 6已知：如图所示，BC为 O的直径，AD BC，垂足为 D，弦 BF和 AD交于 E，且 AE=BE（1）试猜想：AB与AF有何大小关系？并证明你的猜想；（2）若 BD、CD的长是关于x 的方程 x2-kx+16=0 的两个根，求BF的长；（3）在（2）的条件下，若k 为整数，且满足532(12),13713.22kkkk，求 sin2 A的值EOBCADF考前热身训练文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P21要用圆形铁片截出边长为4cm的正方形铁片，求选用的圆形铁片的直径的最小值2圆内两条弦AB和 CD相交于 P点，AB长为 7，AB把 CD分成两部分的线段长为2 和 6，?求 AP的长3如图，PA切 O于点 A，PBC交 O于 B、C，若 PB、PC的长是关于x 的方程 x2-（m-?2）x+（m+2）=0的两个根，且BC=4，求 m的值及 PA的长OBCAP4如图，D是 ABC的边 AC上一点，CD=2AD，AE BC，交 BC于点 E，若 BD=8，sin CBD=34，求 AE的长文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P25如图，在ABC中，CAD=B，若 AD=7，AB=8，AC=6，求 DC的长6已知，如图，以ABC的边 BC为直径的半圆交AB于 D，交 AC于 E，过 E点作 EFBC，?垂足为 F，且 BF：FC=5：1，AB=8，AE=2，求 EC的长EOBCADF答案:中考样题看台1解：3（x+1x）2-5（x+1x）-8=0，x+1x=83或 x+1x=-1，BCAD2xEBCAxDF文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2由 x+1x=83得 x=473 x+1x=-1 得 x2+x+1=0 无解tan ADC=473，在 RtABC中，AC=tan30BC=3在 RtADC中，CD=tanACADC=4 3213CD1，CD=4 32132 PDC=36 3（1）证明：连结AC，ACCE，CEA=CAE CEA=CBA，CBA=CAE，?AB是直径，ACB=90，CP AB，CBA=ACP，CAE=ACP，AD=CD（2）解：ACB=90，CAE=ACP，DCF=CFD，AD=CD=DF=54，ECB=?DAP，tan ECB=34，tan DAP=DPPA=34，PD2+PA2=DA2，DP=34，PA=1，CP=2，ACB=90，CP AB，APC CPB，APPCPCPB，PB=4 4（1）要使方程有两个实数根，必须0，即-（k+1）2-4（14k2+1）0，文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2化简得：2k-3 0，解之得：k32（2）2222(5)1114ababkabk解之得：k1=2，k2=-6 由（1）可知，k=-6 时，方程无实数根，所以，只能取k=25（1）连结 OC，证 OCP=90 即可（2）B=30，A=BCP=60，BCP=CGP=60，CPG 是正三角形PG=CP=43，PC切 O于 CPC2=PD PE=（43）2=48，又 BC=63，AB=6，FD=33，EG=3，PD=23，PD+PE=23+83=103以 PD、PE为两根的一元二次方程为x2-48x+103=0（3）当 G为 BC中点，OG BC，OG AC或 BOG=BAC 时，结论BG2=BFBO成立?要让此结论成立，只证明BFG BGO 即可，凡是能使BFG BGO 的条件都可以6可以猜想到ABAF证明：延工AD交 O于点 GBC是 O的直径，AD BC，ABBG AE=BE，ABE=BAE，AFBG，ABAF（2）ABBGAF，BFAG，BF=AG AD BC，BC是 O直径，AG=2AD，BF=2AD，文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L5H6D8R8 ZY5S6B4E1P2文档编码:CV9S1M10Z4T4 HD8L