鄂爾多斯專版2020中考數(shù)學(xué)復(fù)習(xí)方案第一單元數(shù)與式課時(shí)訓(xùn)練03分式試題

PAGEPAGE1課時(shí)訓(xùn)練(三)分式(限時(shí):30分鐘)|夯實(shí)基礎(chǔ)|1.計(jì)算a2b3·b2A.a5b5 B.a4b5 C.ab5 D.a5b62.化簡(jiǎn)x2x-1A.x+1 B.x-1 C.x2-1 D.x3.[2017·陜西]化簡(jiǎn)xx-y-A.1 B.x2C.x-yx+y D4.[2019·北京]如果m+n=1,那么代數(shù)式2m+nm2-mn+1m·(m2A.-3 B.-1 C.1 D.35.[2019·臨沂]計(jì)算a2a-1-a-1A.-1a-1 C.-2a-1a-1 6.[2019·泰州]若分式12x-1有意義,則7.[2018·濱州]若分式x2-9x-3的值為8.[2018·沈陽]化簡(jiǎn):2aa9.[2017·金華]若ab=23,10.[2019·包頭]化簡(jiǎn):1-a-1a+211.[2019·連云港]化簡(jiǎn)mm2-4÷1+12.[2019·蘇州]先化簡(jiǎn),再求值:x-3x2+6x+9÷1-6x13.[2019·本溪]先化簡(jiǎn),再求值:a2-4a2-4a+4-12-a14.[2017·赤峰]先化簡(jiǎn),再求值:a-6a2-4-3a+2÷aa-2,其中a=201715.[2018·德州]先化簡(jiǎn),再求值:x-3x2-1÷x-3x2+2x|能力提升|16.[2017·樂山]若a2-ab=0(b≠0),則aa+b等于A.0 B.1C.0或12 D.1或17.[2018·孝感]已知x+y=43,x-y=3,則式子x-y+4xyx-yx+y-4xyx+A.48 B.123C.16 D.1218.[2019·濱州]觀察下列一組數(shù):a1=13,a2=35,a3=69,a4=10它們是按一定規(guī)律排列的,請(qǐng)利用其中規(guī)律,寫出第n個(gè)數(shù)an=.(用含n的式子表示)

19.[2017·鎮(zhèn)江]已知實(shí)數(shù)m滿足m2-3m+1=0,則代數(shù)式m2+19m2+220.[2017·杭州]若m-3m-1·m=21.若x+1x=3,求x

【參考答案】1.A2.A3.B4.D[解析]2m+nm2-mn+1m·(m2-n2)=2m+nm(m-n)+m-nm(m-n)·(m+n)(m-故選D.5.B[解析]原式=a2a-1-(故選B.6.x≠127.-38.9.53[解析]解法1:利用比例的基本性質(zhì)“兩內(nèi)項(xiàng)積等于兩外項(xiàng)積”求解.∵ab=23,∴3a=2b,a=23b,∴a+bb=23b+bb=53bb=53.解法2:設(shè)參數(shù)法求解10.-111.解:原式=m(m=m(m=m=1m12.解:原式=x-3(當(dāng)x=2-3時(shí),原式=1213.解:a2-4a2-4a+4-=a+2a-2+1a∵a2+3a-2=0,∴a2+3a=2,∴原式=22=114.解:原式=a-6(a=a-6=-2a=-2∵a=20170+-15-1+27tan30°=1-5+33×∴原式=-2-15.解:原式=x-3(解不等式組5x-3>3(∵x是整數(shù),∴x=4.將x=4代入得,原式=1416.C[解析]∵a2-ab=0(b≠0),∴a(a-b)=0.∴a=0或a-b=0,即a=0或a=b.∴aa+b=017.D[解析]x-y+4xyx-yx+y-4xyx+y=(x-y)2+4xyx-y當(dāng)x+y=43,x-y=3時(shí),原式=43×318.n(n+1)2(2n+1)[解析]這組分?jǐn)?shù)的分子分別為1,3=2+1,6=3+2+1,10=4+3+2+1,15=5+4+3+2+1,…,則第n個(gè)數(shù)的分子為n(n+1)2;分母分別為3=2+1,5=22+1,9=23+1,17=24+1,33=25+1,…,則第19.9[解析]由m2-3m+1=0,可得m2=3m-1.將m2=3m-1代入m2+19m2+2,得3m-1+193m-1+2=3m