山西專用2019中考數(shù)學(xué)復(fù)習(xí)數(shù)與式第3講分式習(xí)題.doc

第3講分式基礎(chǔ)滿分考場零失誤類型一分式及其相關(guān)概念1.(2018金華)(3分)若分式x-3x+3的值為0,則x的值是(A)A.3B.-3C.3或-3D.02.(2018湖北黃岡)(3分)函數(shù)y=x+1x-1中自變量x的取值范圍是(A)A.x-1且x1B.x-1C.x1D.-1x1,8-2x2.預(yù)測(cè)猜押把脈新中考 14.(2019原創(chuàng)預(yù)測(cè))(3分)下列運(yùn)算正確的是(A)A.-x-y-x+y=x-yx+yB.a2-b2(a-b)2=a-ba+bC.a2-b2(a-b)2=a+ba-bD.x-11-x2=1x+115.(2019改編預(yù)測(cè))(5分)化簡求值:1+2a-1a2+2a+1a-1,從-2a2中選一個(gè)合適的值代入求解.16.(2019原創(chuàng)預(yù)測(cè))(5分)先化簡,再求值:x+2x2-2x8xx-2-2+x,其中x=2-1.17.(2019原創(chuàng)預(yù)測(cè))(5分)先化簡,再求值:1-1a-2a2-6a+92a-4,其中a=4cos 30+3tan 45.答案精解精析基礎(chǔ)滿分1.A2.A3.C4.B5.答案1x-16.解析原式=(x-1)2x(x-1)(x+1)x-1x=(x-1)2x(x-1)(x+1)xx-1=1x+1.7.解析原式=x+1-1x+1(x+1)(x-1)x=xx+1(x+1)(x-1)x=x-1.8.解析原式=x2-1-x(x-2)x(x+1)(x+1)22x-1=2x-1x(x+1)(x+1)22x-1=x+1x,當(dāng)x=-2 018時(shí),原式=2 0172 018.9.解析x-3x-4x-1x-2x-1=x2-xx-1-3x-4x-1x-1x-2=x2-4x+4x-1x-1x-2=(x-2)2x-1x-1x-2=x-2.當(dāng)x=12時(shí),原式=12-2=-32.能力升級(jí)10.答案3x-y11.解析原式=x+2x-3+x2-x-6x-3x-3(x-2)2=x2-4x-3x-3(x-2)2=(x+2)(x-2)x-3x-3(x-2)2=x+2x-2.12.解析原式=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(dāng)a=-12時(shí),原式=-4.13.解析原式=x2x2-11+x-1x-1=x2x2-1x-1x=xx+1,解不等式組得2x3,x為整數(shù),x=3,當(dāng)x=3時(shí),原式=33+1=34.預(yù)測(cè)猜押14.C15.解析原式=a-1+2a-1a-1(a+1)2=a+1a-1a-1(a+1)2=1a+1.由題可得a1,從-2a2中選取除1外的值均可代入求值,如取a=0,則原式=1(答案不唯一).16.解析原式=x+2x2-2x8x+(x-2)2x-2=x+2x(x-2)8x+x2-4x+4x-2=x+2x(x-2)x-2(x+2)2=1x(x+2),當(dāng)x=2-1時(shí),原式=1(2-1)(2+1)=1.17.解析原式=a-2-1a-2(a-3)22(a-2)=a-3