【20套试卷合集】四川省广安市名校2019-2020学年数学九上期中模拟试卷

A．(x﹣6)²=﹣4+36C．(x﹣3)²=﹣4+9B．(x﹣6)²=4+36D．(x﹣3)²=4+9∠BAB′=(A．30°B．35°C．40°D．50°DBBDBCBC9(2,-4)-1<<5200(1-x)2=128x2=-1(2)Y1=（-3+Y2=（-3-；-----------3分，--------------5分，------------6分∴k=-3。------------8分A2(-1,-4)B2（-2,-2），∵△ABP的面积为10，∴AB•|n|=10，解得：n=±5，当n=5时，m2+2m﹣3=5，解得：m=﹣4或2，∴P（﹣4，5）2，5）则w=（x-20）（-10x+500）=-10x2+700x-10000；（2）w=-10x2+700x-10000=-10（x-35）2+2250．将x=4，y=0代入y=-x2+bx+c得0=-16+4b+2，解得b=7/22+7/2t+2)=-t2+4t-------------------6，从而MN=-t2+7/2t+2-(-1/2t+2)=-(t-2)ABx()AOMABxAOMA．18B．24C．27D．30A．2+23=35B．8=42C．27÷3=3D．25=±5B.10A．43B．83D．25a2cb2ac2=aFEO(1)932748（2）(21231)6P，顶点坐标2a4aA．50°B．80°C．100°D．200°M6．B；7．B；8．C；9．A；61664【解答】解：原式=2+﹣+=+2．A．4B．2C．21D．20xx28ABAC⌒BA13．解：18(②抛物线经过点(-3,8)；或AM•MM13．10．2.11．•M∴S△ABC=5×6×=15，∴S△ABE=S△AEF+S△BEF=××3=FMMS四边形MPA=SAMB1A．3B．5D．2543=50°，_°．AB24开方得t﹣=±，A．0.5B．4aD．10C．12B．211．在二次根式6、8、12、27中，与18是同类二次根式的是GB8（1）(1220)(35)（2）(221)(122)(31)218.1.5．=335（2）原式=(221)(221)(31)22232m2m13m52AB∴AC=AO=CO∴»»ACAP∴∠ACP=∠ABC=30°A.54B.30C.48D.18A.4333=1B.23=51=2D.322=523x(5x2)2x24xx295A：23B：23C：842D：4215.计算：11(=4-3∴x=-1±22,x2=12.－3x＋53=0，－3a＋53=23,A．62C．32∴b2－4ac＝(－8)2－4×4×3＝16＞02×4MM(x－5)2＝16························2分x－5＝4或x－5＝－4······················3分x1＝9或x2＝1．·······················4分(x－2)(x＋1)＝0··························6分x－2＝0或x＋1＝0··························7分x1＝2或x2＝－1．··························8分(x－3)(3x－2)＝0··························2分x－3＝0或3x－2＝0························3分x1＝3或x2＝3．·························4分∴b2－4ac＝(－3)2－4×2×1＝1＞0．···················5分2×2•······················6分∴x1＝1或x2＝2．····························8分解：(1)△AB1C1，△A1B2C2如图所示；·····················4分(2)B1(－2，－3)，C2(3，1)；························6分解得a＝－1．······························3分∴二次函数的关系式是y＝－(x＋1)2＋4；··················4分∴图象与y轴的交点坐标为(0，3)．·····················5分可以得出平行于墙的一边的长为(25－2x＋1)m，················1分由题意得x(25－2x＋1)＝80，·······················3分解得：x1＝5，x2＝8，···························4分当x＝5时，26－2x＝16＞12(舍去)，····················5分当x＝8时，26－2x＝10＜12．·······················6分解：(1)y＝(x2－2x＋1)－4·························1分＝(x－1)2－4；···························3分(2)令y＝0，得x2－2x－3＝0，·······················4分解得x1＝3，x2＝－1，···························5分函数图象与x轴的交点坐标为(3，0)，(－1，0)．···············6分(1)将x＝1代入方程得：m＋1＋1＝0，····················2分解得：m＝－2；······························3分∴△＝b2－4ac＝1－4m＞0，且m解得：m＜4且m得：9＋3b＋c＝0，解得：c＝－3，····················2分∴抛物线的解析式为y＝x2－2x－3．·····················3分∴顶点坐标为(1，－4)．··························4分(2)由图可得当0＜x＜3时，－4设P(x，y)，则S△PAB＝2AB•|y|＝2|y|＝10，∴|y|＝5，∴y＝±5．········6分综上所述，P点坐标为(－2，5)或(4，5)．··················7分设y＝x2，则原方程变为：y2－2y－8＝0．··················1分解得，y1＝－2，y2＝4，··························2分当y＝－2时，x2＝－2，此方程无实数解；··················3分当y＝4时，x2＝4，解得x1＝－2，x2＝2，··················4分3．····················∴过A、B、C三点的抛物线的表达式为y＝－4x2＋2x＋4．···········3分当BQ∥CP且BQ＝CP时，四边形BCPQ为平行四边形．·············4分∴6－t＝t．解得：t＝3．·························5分∴MN＝yM－yN＝(－4m2＋2m＋4)－(－2m＋4)＝－4m2＋2m．············7分MEFABFD1PAAB∴AD=AB,∠D=∠ABC(2)∵FF∵DE=CE,∴∠CDE=∠ECD,又∵∠CDE+∠BED=∠ABC=∠ACD=∠ECD+∠GCE，∴∠BED=∠GCE∵DE=CE,∴∠CDE=∠ECD,又∵∠CDE-∠BED=∠ABC=∠ACD=∠ECD-∠GCE，∴∠BED=∠GCE又∵AF=BE∴∠CAD1=45°，∠BAD1=135°D1PAABAB3故此可M的值是．则DNBNCBBN21616∵∠P