1.3分式過關(guān)演練(30分鐘80分)1.(xx甘肅白銀)若分式x2-4x的值為0,則x的值是(A)A.2或-2B.2C.-2D.0【解析】分式x2-4x的值為0,x2-4=0,解得x=2或-2.2.(xx山東淄博)化簡a2a-1-1-2a1-a的結(jié)果為(B)A.a+1a-1B.a-1C.aD.1【解析】原式=a2a-1+1-2aa-1=(a-1)2a-1=a-1.3.(xx四川南充)已知1x-1y=3,則代數(shù)式2x+3xy-2yx-xy-y的值是(D)A.-72B.-112C.92D.34【解析】1x-1y=3,y-xxy=3,x-y=-3xy,則原式=2(x-y)+3xy(x-y)-xy=-6xy+3xy-3xy-xy=-3xy-4xy=34.4.(xx江蘇蘇州)計(jì)算1+1xx2+2x+1x的結(jié)果是(B)A.x+1B.1x+1C.xx+1D.x+1x【解析】原式=xx+1x(x+1)2x=x+1xx(x+1)2=1x+1.5.若分式2aba-b中的a,b的值同時(shí)擴(kuò)大到原來的3倍,則分式的值(B)A.不變B.是原來的3倍C.是原來的6倍D.是原來的9倍【解析】若分式2aba-b中的a,b的值同時(shí)擴(kuò)大到原來的3倍,即23a3b3a-3b=6aba-b=32aba-b,所以分式的值是原來的3倍.6.下列分式是最簡分式的是(B)A.1-xx-1B.2xx2+1C.x-1x2-1D.42x【解析】1-xx-1=-x-1x-1=-1,故A錯(cuò)誤;x-1x2-1=x-1(x+1)(x-1)=1x+1,故C錯(cuò)誤;42x=2x,故D錯(cuò)誤.7.(xx北京)如果a-b=23,那么代數(shù)式a2+b22a-baa-b的值為(A)A.3B.23C.33D.43【解析】原式=a2+b22a-2ab2aaa-b=(a-b)22aaa-b=a-b2,當(dāng)a-b=23時(shí),原式=232=3.8.已知14m2+14n2=n-m-2,則1m-1n的值等于(C)A.1B.0C.-1D.-14【解析】由14m2+14n2=n-m-2,得(m+2)2+(n-2)2=0,則m=-2,n=2,1m-1n=-12-12=-1.9.(xx沈陽)化簡:2aa2-4-1a-2=1a+2.【解析】原式=2a(a+2)(a-2)-a+2(a+2)(a-2)=a-2(a+2)(a-2)=1a+2.10.(xx黑龍江大慶)已知3x-4(x-1)(x-2)=Ax-1+Bx-2,則實(shí)數(shù)A=1.【解析】Ax-1+Bx-2=A(x-2)(x-1)(x-2)+B(x-1)(x-1)(x-2)=(A+B)x-(2A+B)(x-1)(x-2),3x-4(x-1)(x-2)=Ax-1+Bx-2,A+B=3,2A+B=4,解得A=1,B=2.11.(8分)先化簡,再求值:1a-2+1a2-13a-6,其中a=tan 60-|-1|.解:a=tan 60-|-1|,a=3-1,原式=a-1a-23(a-2)(a+1)(a-1)=3a+1=3.12.(10分)某學(xué)生化簡分式1x+1+2x2-1出現(xiàn)了錯(cuò)誤,解答過程如下:原式=1(x+1)(x-1)+2(x+1)(x-1)(第一步)=1+2(x+1)(x-1)(第二步)=3x2-1.(第三步)(1)該學(xué)生解答過程是從第步開始出錯(cuò)的,其錯(cuò)誤原因是;(2)請寫出此題正確的解答過程.解:(1)一;分式的基本性質(zhì)用錯(cuò).(2)原式=x-1(x+1)(x-1)+2(x+1)(x-1)=x+1(x+1)(x-1)=1x-1.13.(10分)請你先化簡a2a+2-a+24aa2-4,再從-2,2,2中選擇一個(gè)合適的數(shù)代入求值.解:a2a+2-a+24aa2-4=a2a+2-(a-2)(a+2)a+2(a+2)(a-2)4a=4a+2(a+2)(a-2)4a=a-2a,為使分式有意義,a不能取2,所以a=2,當(dāng)a=2時(shí),原式=2-22=1-2.14.(10分)先化簡,再求值:a-1a2-4a+4-a+2a2-2a4a-1,其中a是滿足不等組7-a>2,2a>3的整數(shù)解.解:a-1a2-4a+4-a+2a2-2a4a-1=a(a-1)-(a+2)(a-2)a(a-2)2a4-a=4-aa(a-2)2a4-a=1(a-2)2,解不等式組得32<a<5,a=2,3,4,原式中a0,2,4,a=3,當(dāng)a=3時(shí),原式=1(3-2)2=1.名師預(yù)測1.若分式x-4x+4無意義,則x的值是(B)A.x=4B.x=-4C.x=0D.x-4【解析】分式x-4x+4無意義,x+4=0,解得x=-4.2.下列分式:a33x2,x-yx2+y2,m2+n2m2-n2,m+1m2-1,a2-2ab+b2a2-2ab-b2,其中最簡分式有(C)A.2個(gè)B.3個(gè)C.4個(gè)D.5個(gè)【解析】a33x2,x-yx2+y2,m2+n2m2-n2,a2-2ab+b2a2-2ab-b2這4個(gè)分式是最簡分式;而m+1m2-1=m+1(m+1)(m-1)=1m-1.故最簡分式有4個(gè).3.已知x2-3x-4=0,則代數(shù)式xx2-x-4的值是(D)A.3B.2C.13D.12【解析】將已知等式整理為x-4x=3,則xx2-x-4=1x-4x-1=13-1=12.4.如果a+b=2,那么代數(shù)式a-b2aa-ba的值是2.【解析】當(dāng)a+b=2時(shí),原式=a2-b2aaa-b=(a-b)(a+b)aaa-b=a+b=2.5.化簡:1+1x-1x2+xx2-2x+1=x-1x+1.【解析】1+1x-1x2+xx2-2x+1=x-1+1x-1(x-1)2x(x+1)=xx-1(x-1)2x(x+1)=x-1x+1.6.先化簡,再求值:1-1x+1xx2-1,其中x=2+1.解:當(dāng)x=2+1時(shí),原式=xx+1(x+1)(x-1)x=x-1=2.7.先化簡xx-2-4x2-2xx+2x2-x,然后從-2,-1,0,1,2中選取一個(gè)你喜歡的值代入求值.解:原式=xx-2-4x(x-2)x+2x(x-1)=x2-4x(x-2)x(x-1)x+2=(x+2)(x-2)x(x-2)x(x-1)x+2=x-1.x=-2,0,1,2能使分母為0,無意義,x只能取-1,當(dāng)x=-1時(shí),原式=-1-1=-2.8.先化簡,再求值:x2-1x2-4-3x-2-x-2,其中x是滿足不等式-12(x-1)12的非負(fù)整數(shù)解.解:-12(x-1)12,x0,即非負(fù)整數(shù)解為0,x=0,原式=(x+1)(x-1)(x+2)(x-2)-3x-2-(x+2)(x-2)x-2=(x+1)(x-1)(x+2)(x-2)x-2(1-x)(1+x)=-1x+2=-12.9.觀察下列等式:第一個(gè)等式:a1=21+32+222=12+1-122+1;第二個(gè)等式:a2=221+322+2(22)2=122+1-123+1;第三個(gè)等式:a3=231+323+2(23)2=123+1-124+1;第四個(gè)等式:a4=241+324+2(24)2=124+1-125+1;按上述規(guī)律,回答下列問題:(1)用含n的代數(shù)式表示第n個(gè)等式an;(2)求a1+a2+a3+a4+a5+a6的最簡結(jié)果;(3)計(jì)算:a1+a2+an.解:(1)an=2n1+32n+2(2n)2=12n+1-12n+1+1.(2)a1+a2+a3+a4+a5+a6=12+1-122+1+122+1-123+1+123+1-124+1+124+1-125+1+125+1-126+1+126+1-127+1=12+1-127+1=1443.(3)a1+a2+an=12+1-122+1+122+1-123+1+12n+1-12n+1+1=12+1-12n+1+1=2n+1-23(2n+1+1).