（福建专版）中考数学复习提分专练01数与式的运算.docx

提分专练(一)数与式的运算|类型1|实数运算1.2019连云港计算:(-1)2+4+13-1.2.2019南平适应性检测计算:2sin30-(-2)0+|3-1|+12-1.3.2019龙岩质检(1)计算:112+123+134+145+156;(2)求证:13113+124+135+14645.|类型2|整式运算4.2019常州如果a-b-2=0,那么代数式1+2a-2b的值是.5.2019常德若x2+x=1,则3x4+3x3+3x+1的值为.6.2019宁德质检先化简,再求值:(x-3)2+x(2+x)-9,其中x=-3.7.2018三明质检先化简,再求值:x(x+2y)-(x+1)2+2x,其中x=3+1,y=3-1.8.若x+y=3,且(x+3)(y+3)=20.(1)求xy的值;(2)求x2+3xy+y2的值.|类型3|分式化简求值9.2019厦门质检化简并求值:2a2-4a2-1a2+2aa2,其中a=2.10.2018莆田质检先化简,再求值:aa2+2a+11-1a+1,其中a=3-1.11.2018龙岩质检先化简,再求值:x-3x2-1x2+2x+1x-3-1,其中x=2+1.12.2019漳州质检先化简,再求值:a2+b2a-2ba2-b2a,其中a=2-1,b=1.13.2019龙岩质检先化简,再求值:x-21+2x+x2x-3xx+1,其中x=13.14.2019本溪先化简,再求值:a2-4a2-4a+4-12-a2a2-2a.其中a满足a2+3a-2=0.【参考答案】1.解:原式=-2+2+3=3.2.解:原式=212-1+3-1+2=1-1+3-1+2=3+1.3.解:(1)原式=1-12+12-13+13-14+14-15+15-16=1-16=56.(2)证明:方法一:113+124+135+146=121-13+1212-14+1213-15+1214-16=121-13+12-14+13-15+14-16=1730.13=1030,45=2430,13=103017302430=45,即原式得证.方法二:123+134+145+156113+124+135+146112+123+134+145,12-13+13-14+14-15+15-16113+124+135+1461-12+12-13+13-14+14-15,13113+124+135+14645,即原式得证.4.55.4解析3x4+3x3+3x+1=3x2(x2+x)+3x+1=3x2+3x+1=3(x2+x)+1=4.6.解:原式=x2-6x+9+2x+x2-9=2x2-4x.当x=-3时,原式=2(-3)2-4(-3)=6+43.7.解:原式=x2+2xy-(x2+2x+1)+2x=x2+2xy-x2-2x-1+2x=2xy-1.当x=3+1,y=3-1时,原式=2(3+1)(3-1)-1=2(3-1)-1=3.8.解:(1)(x+3)(y+3)=20,xy+3x+3y+9=20,即xy+3(x+y)=11.将x+y=3代入得xy+9=11,xy=2.(2)当xy=2,x+y=3时,原式=(x+y)2+xy=32+2=9+2=11.9.解:2a2-4a2-1a2+2aa2=2a2-4-a2a2a2a2+2a=(a+2)(a-2)a2a2a(a+2)=a-2a.当a=2时,原式=2-22=1-2.10.解:原式=a(a+1)2a+1-1a+1=a(a+1)2a+1a=1a+1.当a=3-1时,原式=13-1+1=13=33.11.解:原式=x-3(x+1)(x-1)(x+1)2x-3-1=x+1x-1-x-1x-1=2x-1.当x=2+1时,原式=22+1-1=22=2.12.解:原式=(a-b)2aa(a-b)(a+b)=a-ba+b,当a=2-1,b=1时,原式=2-22=1-2.13.解:原式=x-2(x+1)2x2+xx+1-3xx+1=x-2(x+1)2x+1x(x-2)=1x(x+1).当x=13时,原式=11343=94.14.解:a2-4a2-4a+4-12-a2a2-2a=(a-2)(a+2)(a