浙江省2019年中考數(shù)學(xué) 第一單元 數(shù)與式 課時(shí)訓(xùn)練03 分式練習(xí) （新版）浙教版.doc

課時(shí)訓(xùn)練(三)分式夯實(shí)基礎(chǔ)1.xx綿陽(yáng) 等式x-3x+1=x-3x+1成立的x的取值范圍在數(shù)軸上可表示為()圖K3-12.xx宜昌 計(jì)算(x+y)2-(x-y)24xy的結(jié)果為()A.1B.12C.14D.03.xx南充 已知1x-1y=3,則代數(shù)式2x+3xy-2yx-xy-y的值是()A.-72B.-112 C.92D.344.下列運(yùn)算結(jié)果為x-1的是()A.1-1xB.x2-1xxx+1C.x+1x1x-1D.x2+2x+1x+15.xx衢州 當(dāng)x=6時(shí),分式51-x的值等于.6.觀察下列一組數(shù):32,1,710,917,1126,它們是按一定規(guī)律排列的,那么這組數(shù)的第n個(gè)數(shù)是(n為正整數(shù)).7.xx包頭 化簡(jiǎn):a2-1a2(1a -1)a=.8.某班a名同學(xué)參加植樹(shù)活動(dòng),其中男生b名(b<a),若只由男生完成,每人需植樹(shù)15棵,若只由女生完成,則每人需植樹(shù)棵.9.觀察規(guī)律并填空:1-122=1232=34;(1-122)(1-132)=12322343=1243=23;(1-122)(1-132)(1-142)=123223433454=1254=58;(1-122)(1-132)(1-142)(1-152)=1232234334544565=1265=35;(1-122)(1-132)(1-142)(1-1n2)=(用含n的代數(shù)式表示,n是正整數(shù),且n2).10.(1)化簡(jiǎn):4a+4b5ab15a2ba2-b2.(2)xx鹽城 先化簡(jiǎn),再求值:(1- 1x+1)xx2-1,其中x=2+1.11.xx曲靖 先化簡(jiǎn),再求值(1a-b-ba2-b2)a2-aba2-2ab+b2,其中a,b滿足a+b-12=0.12.已知A=x2+2x+1x2-1-xx-1.(1)化簡(jiǎn)A;(2)當(dāng)x滿足不等式組x-10,x-3<0且為整數(shù)時(shí),求A的值.13.化簡(jiǎn):aa2-4a+2a2-3a-12-a,并求值.其中a與2,3構(gòu)成ABC的三邊長(zhǎng),且a為整數(shù).拓展提升14.xx達(dá)州 化簡(jiǎn)代數(shù)式:( 3xx-1 - xx+1)xx2-1,再?gòu)牟坏仁浇Mx-2(x-1)1,6x+10>3x+1的解集中取一個(gè)合適的整數(shù)值代入,求出代數(shù)式的值.15.對(duì)于任意的實(shí)數(shù)x,記f(x)=2x2x+1.例如:f(1)=2121+1=23,f(-2)=2-22-2+1=15.(1)求f(2),f(-3)的值;(2)試猜想f(x)+f(-x)的值,并說(shuō)明理由;(3)計(jì)算:f(-xx)+f(-xx)+f(-1)+f(0)+f(1)+f(xx)+f(xx).參考答案1.B解析 由等式x-3x+1=x-3x+1成立,可得x-30,x+1>0,解得x3.故選B.2.A解析 根據(jù)整式的運(yùn)算法則及分式的基本性質(zhì)化簡(jiǎn),原式=x2+y2+2xy-x2-y2+2xy4xy=4xy4xy=1.3.D解析 1x-1y=3,y-x=3xy,x-y=-3xy,原式=2(x-y)+3xy(x-y)-xy=-6xy+3xy-3xy-xy=-3xy-4xy=34.4.B5.-16.2n+1n2+17.-a-18.15ba-b9.n+12n10.解:(1)4a+4b5ab15a2ba2-b2=4(a+b)5ab15a2b(a+b)(a-b)=12aa-b.(2)原式=x+1-1x+1x2-1x=xx+1(x+1)(x-1)x=x-1.當(dāng)x=2+1時(shí),原式=2+1-1=2.11.解:( 1a-b-ba2-b2)a2-aba2-2ab+b2=a+b(a+b)(a-b)-b(a+b)(a-b)(a-b)2a(a-b)=a+b-b(a+b)(a-b)a-ba=aa+b1a=1a+b.由于a,b滿足a+b-12=0,所以a+b=12,因此原式化簡(jiǎn)后的式子:1a+b=112=2.12.解:(1)A=(x+1)2(x+1)(x-1)-xx-1=x+1x-1-xx-1=x+1-xx-1=1x-1.(2)解不等式組,得1x<3.x為整數(shù),x=1或x=2.A=1x-1,x1.當(dāng)x=2時(shí),A=1x-1=12-1=1.13.解:原式=a(a+2)(a-2)a+2a(a-3)+1a-2=1(a-2)(a-3)+a-3(a-2)(a-3)=a-2(a-2)(a-3)=1a-3.a與2,3構(gòu)成ABC的三邊長(zhǎng),3-2<a<3+2,即1<a<5,a為整數(shù),a的值為2或3或4.當(dāng)a=2時(shí),分母2-a=0,舍去;當(dāng)a=3時(shí),分母a-3=0,舍去;故a的值只能為4.當(dāng)a=4時(shí),原式=14-3=1.14.解:解不等式,得x1,解不等式,得x>-3,不等式組x-2(x-1)1,6x+10>3x+1的解集為-3<x1.(3xx-1-xx+1)xx2-1=3x(x+1)-x(x-1)(x-1)(x+1)x2-1x=3x(x+1)-x(x-1)(x-1)(x+1)(x-1)(x+1)x=3(x+1)-(x-1)=3x+3-x+1=2x+4.x0,x1,當(dāng)x取-2時(shí),原式=2(-2)+4=0.15.解:(1)f(2)=2222+1=45,f(-3)=2-32-3+1=19.(2)猜想:f(x)+f(-x)=1.理由:f(x)+f(-x)=2x2x+1+2-x2-x+1=2x+12x+1=1.(3)原式=xx1+f(0)=xx+12=40332.