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1、[XD]KZDA(2KHTK3[CS%0,0,0,0][CSX%0,0,0,0]KZD)3KZDB(3KHTSS2KCS3HCSX3KZD)3KZDC(3KJYU(KHTKU[[KG*?!]EWTBZU[CS%0,0,0,0][CSX%0,0,0,0]Kzd)3Kzdd(3KhtssUKkg*23KcsuKcsxH)KwtbxUKzd)UKBW(D(Z3mm,・11mm,・13mm)3gXC正封.tifUKBW)UKBW(D(X5mm,,)MD2mm3[KG72mm][HT5]—105—[HT]KKG147.
2、5mm3[HT5]—106—[HT]KBW)3KBW(S(X5mm,,)MD2mm3[KG72mm][HT5]—107—[HT]KKG147.5mm3[HT5]—108—[HT]KBW)!][WTBX][FL(K20mm]KJZ3KHT3DBS3九年级人教版数学(上)KHT3KHS2HKJZ2KHT3CQU第二^一章测评卷KHTUKHTUKjz3Hht5k3(时间：120分钟=满分：loo分)Kht3KjzUKht53Kfk(W3Kbg(2KBHDFG2*2,FK30mm,K28mm。3,K30mmF2题=号
3、o-o-n三o总分KbhdS得=分Kbg)f3Kfk)HKht2KHJ3.6mmHKHT5H3一、选择题[HTK](每小题4分，共32分)[HTSS]KHT31.[ZK(]下列方程是一元二次方程的是[=C]D[=D]A.x+2y=1======二■■[WB]B.[2x(x-l)=2x2+3=：=======[WB]C.13x+—=4=====：X===[WB]D.x2—2=0[ZK)]3.[ZK(]如果[KF(][SX(]x[]x-6[SX)][KF)]=[SX(][KF(]x[KF)][][KF(]x-6[
4、KF)][SX)]成立，那么x应满足的条件是[=C]D[=D]A.xN6[DW]B.0WxW6[DW]C.x>0[DW]D.x>6[ZK)]4.[ZK(]已知△ABC的三边a,b,c满足a2+b+
5、[KF(]c-l[KF)]-2
6、=10a+2[KG-*2][KF(]b-4[KF)]-22,则AABC为[=C]B[=D]A.等腰三角形[DW2]B.等边三角形C.直角三角形[DW2]A.等腰直角三边形[ZK)]3.[ZK(]下列计算正确的是[=C]B[=D]A.3+[KF(]3[KF)]=3[KF(]3[KF)]
7、[DW2]B.[KF(]27[KF)]4-[KF(]3[KF)]=3C.[KF(]2[KF)]・[KF(]3[KF)]=[KF(]5[KF)][DW2]D.[KF(]4[KF)]=±2[ZK)]4.[ZK(]设a=[KF(]3[KF)卜[KF(]2[KF)],b=2-[KF(]3[KF)],c=[KF(]5[KF)卜2,则a、b、c的大小关系是[=C]A[=叨A.a>b>c[DW]B.a>c>b[DW]C.c>b>a[DW]D.b>c>a[ZK)]5.[ZK(]若x=[KF(]m[KF)]-[KF(]n[KF
8、)],y=[KF(]m[KF)]+[KF(]n[KF)],则xy的值是[=C]D[=D]A.2[KF(]m[KF)][DW]B.2[KF(]n[KF)][DW]C.m+n[DW]D.m-n[ZK)]6.[ZK(]估算[KF(]24[KF)]+3的值是[=C]C[=D]A.在5和6之间[DW2]B.在6和7之间C.在7和8之间[DW2]D.在8和9之间[ZK)]UHTH2二、填空题KHTK2(每小题3分，共18分)KHT37.KZK(3把[KF(][SX(]3[]x[SX)][KF)]化成最简二次根式为[ZZ(
9、Z]=[=A][SX(]l[]x[SX)][KF(]3x[KF)][=B]=[ZZ)].[ZK)38.KZK(2若[KF(]x2-9[KF)]=[KF(]x+3[KF)]・[KF(]x・3[KF)]成立，则x应满足的条件是[ZZ(Z]=[=A]x^3[=B]=[ZZ)].KZK"9.KZK(3计算：[KF(]([KF(]3[KF)卜2)2[KF)]+[KF(]3[KF)]=[ZZ(Z]==[=A]2[=B]==[ZZ)].KZK)》10.KZK(3已知一次函数y=(m-2)x+(3-m)的图象经过第一、二、四
10、象限，化简[KF(]m2・4m+4[KF)]+[KF(]9・6m+m2[KF)]=[ZZ(Z]=[=A]5-2m[=B]=[ZZ)].EZK)311.KZK(2设54-[KF(]11[KF)]的小数部分为a,5-[KF(]ll[KF)]的小数部分为b,则a+b珂ZZ(Z]=[=A]1[=B]=[ZZ)].Kzk)3[TPA160.TIF;Z*2,Y,PZ]12.KZK(3如图，已知圆柱体底面圆的半径
