八年級(jí)上冊(cè)數(shù)學(xué)知識(shí)點(diǎn)提升練習(xí)02-全練版：15.2.2　分式的加減

第十五章分式15.2分式的運(yùn)算15.2.2分式的加減基礎(chǔ)過(guò)關(guān)全練知識(shí)點(diǎn)1分式的加減法法則1.(2022天津中考)計(jì)算a+1a+2+1a+2A.1B.2a+2C.a2.(2021江西中考)計(jì)算a+1a-1a的結(jié)果為(A.1B.-1C.a+2a3.(2020山東臨沂中考)計(jì)算xx?1-yy?1的結(jié)果為A.?x+C.?x?4.計(jì)算a2a+1-a-1的結(jié)果正確的是(A.-1a+1B.1a+15.(2022湖南益陽(yáng)中考)計(jì)算:2aa?1-26.計(jì)算24m4m2?97.計(jì)算:(M8115002)(1)m-1+2m?6m(2)x+2x2?2x8.(2019湖北宜昌中考)已知:x≠y,y=-x+8,求x2x?y知識(shí)點(diǎn)2分式的混合運(yùn)算9.【新獨(dú)家原創(chuàng)】小林在做作業(yè)時(shí)發(fā)現(xiàn)一道題有一部分被墨滴遮蓋了,如圖所示,小林查看了答案,正確結(jié)果為a1?a,則“”是()A.-aB.-bC.aD.b10.(2021內(nèi)蒙古包頭中考)化簡(jiǎn):2mm2?4+11.(2020山東濟(jì)寧中考)已知m+n=-3,則m+nm÷?12.(2022遼寧大連中考)計(jì)算:x2?4x2?4x+4÷13.(2021廣東深圳中考)先化簡(jiǎn),再求值:1x+2+1÷x2+614.(2020四川遂寧中考)先化簡(jiǎn),x2+4x+4x2?4?x能力提升全練15.(2022四川眉山中考,8,★★☆)化簡(jiǎn)4a+2+a-2的結(jié)果是(A.1B.a2a+2C.16.(2021山東濟(jì)南中考,7,★★☆)計(jì)算m2m?1-2m?1A.m+1B.m-1C.m-2D.-m-217.【跨學(xué)科·物理】(2022浙江杭州中考,6,★★☆)照相機(jī)成像應(yīng)用了一個(gè)重要原理,用公式1f=1u+1v(v≠f)表示,其中f表示照相機(jī)鏡頭的焦距,u表示物體到鏡頭的距離,v表示膠片(像)到鏡頭的距離.已知f,v,則u=(A.fvf?vB.f?18.(2022山東威海中考,7,★★☆)試卷上一個(gè)正確的式子1a+b+1a?bA.aa?bB.a?19.(2019北京中考,6,★★☆)如果m+n=1,那么2m+nm2?mn+1m·(mA.-3B.-1C.1D.320.(2022山東菏澤中考,13,★★☆)若a2-2a-15=0,則a?4a?4a21.(2021遼寧沈陽(yáng)中考,13,★★☆)計(jì)算:1x?4?8x2?16·22.(2023山東青島聯(lián)考,16,★★☆)計(jì)算:2x+4x2?623.(2021黑龍江綏化中考,16,★★☆)當(dāng)x=2021+3時(shí),x+3x2?3x24.(2023北京師大附中期末,20,★★☆)計(jì)算:m2?4m+425.(2021湖南郴州中考改編,18,★★☆)先化簡(jiǎn),再求值:a?1a2+a?a26.(2022河南鄭州模擬,18,★★☆)先化簡(jiǎn)mm+3?2mm?3÷mm227.(2022湖南婁底中考,20,★★☆)先化簡(jiǎn),再求值:x+2+4x?2÷x3x2素養(yǎng)探究全練28.【運(yùn)算能力】已知a,b,c為實(shí)數(shù),且a+bab=3,b+c(1)求1a+1b+1(2)求ab+bc(3)分別求a,b,c的值.

答案全解全析基礎(chǔ)過(guò)關(guān)全練1.A原式=a+1+1a+2=a+22.A原式=a+1?1a=aa=13.A原式=x(y?1)(x?1)(y?1)-4.Ca2a+1-a-1=a2a+1-(a5.答案2解析原式=2a?2a?1=2(6.答案2解析原式=24m(2=24m+4m2?127.解析(1)m-1+2m?6m2?9-2m=(m?1)(m+3)=(=m2+2m?3+2?2m?8(2)x+2x2?2=x+2x(x=x+2?x28.解析x2x?y+y2y?x=x2x?y-y2x?y=x2?y2x?y=(x+9.D∵÷a2b?b∴ba+1-=a1?a·b∴=ba+1+aba+1=10.答案1解析原式=2m?(m+2)(m+2)(m?2)11.答案1解析原式=m+n=m+nm·m當(dāng)m+n=-3時(shí),原式=13.故答案為112.解析x2?4x2?4x+4÷x2+2x2x?4-113.解析原式=1+x+2x+2·x+3x2當(dāng)x=-1時(shí),原式=1?1+2=114.解析原式=(x+2)=x+2x?2?x2=-(x+2)(x?3)x?2·x?2x+2∵x≠±2,∴可取x=1,則原式=-1+3=2.(答案不唯一)能力提升全練15.B4a+2+a-2=4a+2+a216.B原式=m2?(2m?1)m?1=m2?217.C∵1f=1u+1v(v≠f),∴1u=1f∴u=fvv?f18.A∵1a+b∴被墨汁遮住部分的式子是1a+=a?b+a+b(19.D原式=2m+n+m?nm(m?n)·(m+n)(m-n)=3mm(m?n)·(m+n)(20.答案15解析原式=a2?4a+4a·=a(a-2)=a2-2a,∵a2-2a-15=0,∴a2-2a=15,∴原式=15.21.答案1解析1x?4?8x=x+4?8(x+4)(x?4)·(x+4)=x?4故答案為1.22.答案2解析2x+4x2?6x=2(x+2)(x?3)23.答案1解析原式=x2?9=x?9x(x?3)當(dāng)x=2021+3時(shí),原式=1(2021+3?3)24.解析m2?4m=(m?2)2=(m?2)2m?1=2?m2+m+1=2?25.解析a?1a=a?1a(a+1)=(a?1)2?a=a2?2a+1?a=a+1a(a+1)(a?1)·當(dāng)a=2時(shí),原式=1226.解析原式=m(m?3)(m+3)(=m[(m=(m-3)-2(m+3)=m-3-2m-6=-m-9,當(dāng)m=-3,0,3時(shí),原式?jīng)]有意義,舍去;當(dāng)m=1時(shí),原式=-1-9=-10.27.解析原式=x2?4=x2x?2·(∵x≠0且x-2≠0,∴x≠0且x≠2,∴x=1,則原式=1?21=-1素養(yǎng)探究全練