第3講 分式（可自主編輯word）

1、第3講分式a組基礎(chǔ)題組一、選擇題1.(2019常州)若代數(shù)式x+1x-3有意義,則實數(shù)x的取值范圍是()a.x=-1 b.x=3 c.x-1 d.x32.(2019新泰模擬)若分式|x|-1(x-2)(x+1)的值為0,則x等于()a.-1 b.-1或2 c.-1或1 d.13.下列分式中,最簡分式是()a.x2-1x2+1 b.x+1x2-1 c.x2-2xy+y2x2-xyd.x2-362x+124.(2019肥城模擬)化簡a+3a-4a-31a-2-1的結(jié)果等于()a.-a-2 b.a-2a-3 c.a+2 d.a-3a-25.下列式子中,錯誤的是()a.nama=nm b.0.5+b0

2、.2a-0.3b=5+10b2a-3b c.nmn+n2=1m+n d.m-m+n=-mm+n6.下列運算結(jié)果為x-1的是()a.1-1x b.x2-1xxx+1 c.x+1x1x-1 d.x2+2x+1x+1二、填空題7.化簡6m2-9+1m+3的結(jié)果是.8.化簡:a2a-3+93-aa+3a=.三、解答題9.(2018濱州)先化簡,再求值:(xy2+x2y)xx2+2xy+y2x2yx2-y2,其中x=0-12-1,y=2sin 45-8.10.(2018德州)先化簡,再求值:x-3x2-1x-3x2+2x+1-1x-1+1,其中x是不等式組5x-33(x+1),12x-19-32x的整數(shù)

3、解.11.(2019 新泰模擬)先化簡,再求值:b2-a2a2-aba+2ab+b2a1a+1b,其中a,b滿足a+3b=-1,3a-2b=8.b組提升組一、選擇題1.化簡m-1mm-1m2的結(jié)果是()a.m b.1m c.m-1 d.1m-12.已知x2-3x-4=0,則代數(shù)式xx2-x-4的值是()a.3b.2c.13d.123.若ab=1,m=11+a+11+b,則2 016m=()a.2 016 b.0 c.1 d.2二、填空題4.(2019內(nèi)江)若1m+1n=2,則分式5m+5n-2mn-m-n的值為.三、解答題5.先化簡,再求值:aa+1-a-1aaa+2-1a2+2a,其中a=-

4、12.6.先化簡,再求值:a-1-2a-1a+1a2-4a+4a+1,其中a=2+3.7.先化簡,再求值:a2+aa2-2a+12a-1-1a,其中a是方程2x2+x-3=0的解.8.先化簡下列各式,再求值:a2-6ab+9b2a2-2ab5b2a-2b-a-2b-1a,其中a,b滿足a+b=5,a-b=3.第3講分式a組基礎(chǔ)題組一、選擇題1.d代數(shù)式x+1x-3有意義,x-30.x3.故選d.2.d分式|x|-1(x-2)(x+1)的值為0,|x|-1=0,x-20,x+10.解得x=1.故d.3.aa.原式為最簡分式,符合題意;b.原式=x+1(x+1)(x-1)=1x-1,不符合題意;c

5、.原式=(x-y)2x(x-y)=x-yx,不符合題意;d.原式=(x+6)(x-6)2(x+6)=x-62,不符合題意.故選a.4.a原式=a2-3aa-3+3a-4a-31a-2-a-2a-2=a2-4a-3-a+3a-2=(a-2)(a+2)a-3-(a-3)a-2=-(a+2)=-a-2.5.d分式的分子與分母同乘(或除以)一個不等于0的整式,分式的值不變.分子、分母、分式本身同時改變兩處的符號,分式的值不變.a.根據(jù)分式的基本性質(zhì),nama=nm成立;b.根據(jù)分式的基本性質(zhì),0.5+b0.2a-0.3b=5+10b2a-3b,成立;c.根據(jù)分式的基本性質(zhì),nmn+n2=1m+n,成立

6、;d.根據(jù)分式的基本性質(zhì),m-m+n=-mm-n,故本選項不成立.故選d.6.ba項的結(jié)果為x-1x,b項的結(jié)果為x-1,c項的結(jié)果為x2-1x,d項的結(jié)果為x+1.二、填空題7.答案1m-3解析6(m+3)(m-3)+m-3(m+3)(m-3)=m+3(m+3)(m-3)=1m-3.8.答案a解析原式=a2a-3-9a-3a+3a=a2-9a-3a+3a=(a+3)aa+3=a.三、解答題9.解析原式=xy(y+x)x(x+y)2(x+y)(x-y)x2y=x-y,當x=1-2=-1,y=2-22=-2時,原式=2-1.10.解析原式=x-3(x+1)(x-1)(x+1)2x-3-1+x-1

7、x-1=x+1x-1-xx-1=1x-1,解不等式組得3x5,其整數(shù)解為x=4,當x=4時,原式=13.11.解析原式=(b-a)(a+b)a(a-b)(a+b)2ab+aab=-1ab,由方程組a+3b=-1,3a-2b=8,解得a=2,b=-1.原式=12.b組提升題組一、選擇題1.a原式=m-1mm2m-1=m.故選a.2.d易知x0,等式整理,得x-4x=3,則原式=1x-4x-1=13-1=12.3.am=11+a+11+b=1+b+1+a1+ab+a+b,把ab=1代入,得m=1,所以2 016m=2 016.二、填空題4.答案-4解析1m+1n=2,可得m+n=2mn,5m+5n

8、-2mn-m-n=5(m+n)-2mn-(m+n)=10mn-2mn-2mn=-4.三、解答題5.解析原式=aa+1-a-1aaa+2-1a(a+2)=aa+1-a-1aa2a(a+2)-1a(a+2)=aa+1-a-1aa2-1a(a+2)=aa+1-a-1aa(a+2)(a+1)(a-1)=aa+1-a+2a+1=-2a+1,當a=-12時,原式=-2-12+1=-4.6.解析原式=(a-1)(a+1)a+1-2a-1a+1a+1(a-2)2=a2-2aa+1a+1(a-2)2=a(a-2)a+1a+1(a-2)2=aa-2,當a=2+3時,原式=2+32+3-2=2+33=23+33.7.解析原式=a(a+1)(a-1)22aa(a-1)-a-1a(a-1)=a(a+1)(a-1)2a+1a(a-1)=a(a+1)(a-1)2a(a-1)a+1=a2a-1,解方程2x2+x-3=0,得x1=1,x2=-32.a-10,即a1,所以a=-32.則原式=94-3