（課標(biāo)通用）甘肅省2019年中考數(shù)學(xué)總復(fù)習(xí)優(yōu)化設(shè)計(jì) 考點(diǎn)強(qiáng)化練3 分式

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1、考點(diǎn)強(qiáng)化練3分式基礎(chǔ)達(dá)標(biāo)一、選擇題1.若分式x-1x+2的值為0,則()A.x=-2B.x=0C.x=1D.x=1或x=-2答案C2.(易錯(cuò)題)下列運(yùn)算錯(cuò)誤的是()A.(a-b)2(b-a)2=1B.-a-ba+b=-1C.0.5a+b0.2a-0.3b=5a+10b2a-3bD.a-ba+b=b-ab+a答案D3.(2018江西)計(jì)算(-a)2ba2的結(jié)果為()A.bB.-bC.abD.ba答案A4.(2017遼寧大連)計(jì)算3x(x-1)2-3(x-1)2的結(jié)果是()A.x(x-1)2B.1x-1C.3x-1D.3x+1答案C二、填空題5.化簡:x2-1x+x+1x=.答案x+16.計(jì)算:a

2、a+b+2ba+baa+2b=.答案aa+b7.如圖所示,圖1是一個(gè)邊長為a的正方形剪去一個(gè)邊長為1的小正方形,圖2是一個(gè)邊長為(a-1)的正方形,記圖1,圖2中陰影部分的面積分別為S1,S2,則S1S2可化簡為.答案a+1a-1三、解答題8.化簡:m2n+mn22m2n2.解m2n+mn22m2n2=mn(m+n)mn2mn=m+n2mn.9.化簡:1+1a-1aa2-2a+1.解原式=aa-1(a-1)2a=a-1.10.(易錯(cuò)題)先化簡,再求值:3x+1-x+1x2+4x+4x+1,其中x=2-2.解原式=3x+1-(x+1)(x-1)x+1x+1(x+2)2=-(x+2)(x-2)x+

3、1x+1(x+2)2=2-xx+2當(dāng)x=2-2時(shí),原式=2-xx+2=2-2+22-2+2=4-22=22-1.11.化簡分式:x2-2xx2-4x+4-3x-2x-3x2-4,并從1,2,3,4這四個(gè)數(shù)中取一個(gè)合適的數(shù)作為x的值代入求值.解x2-2xx2-4x+4-3x-2x-3x2-4=x(x-2)(x-2)2-3x-2x-3x2-4=xx-2-3x-2x-3x2-4=x-3x-2(x+2)(x-2)x-3=x+2,x2-40,x-30,x2且x-2且x3,可取x=1代入,原式=3.(或可取x=4代入,原式=6)導(dǎo)學(xué)號(hào)13814025能力提升一、選擇題1.(2018浙江金華)若分式x-3x

4、+3的值為0,則x的值為()A.3B.-3C.3或-3D.0答案A2.(易錯(cuò)題)化簡a2+2ab+b2a2-b2-ba-b的結(jié)果是()A.aa-bB.ba-bC.aa+bD.ba+b答案A3.如圖所示,將形狀、大小完全相同的“”和線段按照一定規(guī)律擺成下列圖形,第1幅圖形中“”的個(gè)數(shù)為a1,第2幅圖形中“”的個(gè)數(shù)為a2,第3幅圖形中“”的個(gè)數(shù)為a3,以此類推,則1a1+1a2+1a3+1a19的值為()A.2021B.6184C.589840D.421760答案C解析a1=3=13,a2=8=24,a3=15=35,a4=24=46,an=n(n+2);1a1+1a2+1a3+1a19=113+

5、124+135+146+11921=12(1-13+12-14+13-15+14-16+119-121)=12(1+12-120-121)=589840,故選C.4.(2018云南)已知x+1x=6,則x2+1x2=()A.38B.36C.34D.32答案C解析把x+1x=6兩邊平方得:x+1x2=x2+1x2+2=36,則x2+1x2=34,故選C.5.(2018湖北孝感)已知x+y=43,x-y=3,則式子x-y+4xyx-yx+y-4xyx+y的值是()A.48B.123C.16D.12答案D解析x-y+4xyx-yx+y-4xyx+y=(x-y)2+4xyx-y(x+y)2-4xyx+

6、y=(x+y)2x-y(x-y)2x+y=(x+y)(x-y),當(dāng)x+y=43,x-y=3時(shí),原式=433=12,故選D.二、填空題6.a,b互為倒數(shù),代數(shù)式a2+2ab+b2a+b1a+1b的值為.答案1解析a2+2ab+b2a+b1a+1b=(a+b)2a+ba+bab=(a+b)aba+b=ab.又a,b互為倒數(shù),ab=1.7.若實(shí)數(shù)x滿足x2-22x-1=0,則x2+1x2=.答案10解析x2-22x-1=0,x-22-1x=0,x-1x=22,x-1x2=8,即x2-2+1x2=8,x2+1x2=10.三、解答題8.(預(yù)測)先化簡,再求值:xx-3-1x-3x2-1x2-6x+9,其

7、中x滿足2x+4=0.解原式=x-1x-3(x-3)2(x+1)(x-1)=x-3x+1,由2x+4=0,得到x=-2,則原式=x-3x+1=5.9.(2017青海西寧)先化簡,再求值:n2n-m-m-nm2,其中m-n=2.解原式=n2n-m-(m+n)1m2=n2-n2+m2n-m1m2=1n-m,m-n=2,n-m=-2,原式=1n-m=1-2=-22.10.(預(yù)測)先化簡1-1x-1x2-4x+4x2-1,再從不等式2x-16的正整數(shù)解中選一個(gè)適當(dāng)?shù)臄?shù)代入求值.解1-1x-1x2-4x+4x2-1=x-2x-1(x+1)(x-1)(x-2)2=x+1x-2,2x-16,2x7,x72,正整數(shù)解為1,2,3,當(dāng)x=1或x=2時(shí),原式都無意義,x=3,把x=3代入原式得:原式=x+1x-2=3+13-2=4.導(dǎo)學(xué)號(hào)1381402611.先化簡,再求值:(x-1)2x+1-1,其中x為方程x2+3x+2=0的根.解原式=(x-1)2-x-1x+1=(x-1)1-xx+1=(x-1)x+11-x=-x-1.由x為方程x2+3x+2=0的根,解得x=-1或x=-2.當(dāng)x=-1時(shí),原式無意義,所以x=-1舍去;當(dāng)x=-2時(shí),原式=-(-2)-1=2-1=1.6