（福建專版）2020年中考數(shù)學復習 提分專練01 數(shù)與式的運算

提分專練(一)數(shù)與式的運算|類型1|實數(shù)運算1.2019·連云港計算:(-1)×2+4+13-1.2.2019·南平適應(yīng)性檢測計算:2sin30°-(-2)0+|3-1|+12-1.3.2019·龍巖質(zhì)檢(1)計算:11×2+12×3+13×4+14×5+15×6;(2)求證:13<11×3+12×4+13×5+14×6<45.|類型2|整式運算4.2019·常州如果a-b-2=0,那么代數(shù)式1+2a-2b的值是. 5.2019·常德若x2+x=1,則3x4+3x3+3x+1的值為. 6.2019·寧德質(zhì)檢先化簡,再求值:(x-3)2+x(2+x)-9,其中x=-3.7.2018·三明質(zhì)檢先化簡,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.|類型3|分式化簡求值9.2019·廈門質(zhì)檢化簡并求值:2a2-4a2-1÷a2+2aa2,其中a=2.10.2018·莆田質(zhì)檢先化簡,再求值:aa2+2a+1÷1-1a+1,其中a=3-1.11.2018·龍巖質(zhì)檢先化簡,再求值:x-3x2-1·x2+2x+1x-3-1,其中x=2+1.12.2019·漳州質(zhì)檢先化簡,再求值:a2+b2a-2b÷a2-b2a,其中a=2-1,b=1.13.2019·龍巖質(zhì)檢先化簡,再求值:x-21+2x+x2÷x-3xx+1,其中x=13.14.2019·本溪先化簡,再求值:a2-4a2-4a+4-12-a÷2a2-2a.其中a滿足a2+3a-2=0.【參考答案】1.解:原式=-2+2+3=3.2.解:原式=2×12-1+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)證明:方法一:11×3+12×4+13×5+14×6=121-13+1212-14+1213-15+1214-16=121-13+12-14+13-15+14-16=1730.13=1030,45=2430,13=1030<1730<2430=45,即原式得證.方法二:12×3+13×4+14×5+15×6<11×3+12×4+13×5+14×6<11×2+12×3+13×4+14×5,12-13+13-14+14-15+15-16<11×3+12×4+13×5+14×6<1-12+12-13+13-14+14-15,13<11×3+12×4+13×5+14×6<45,即原式得證.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.當x=-3時,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.當x=3+1,y=3-1時,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.將x+y=3代入得xy+9=11,xy=2.(2)當xy=2,x+y=3時,原式=(x+y)2+xy=32+2=9+2=11.9.解:2a2-4a2-1÷a2+2aa2=2a2-4-a2a2·a2a2+2a=(a+2)(a-2)a2·a2a(a+2)=a-2a.當a=2時,原式=2-22=1-2.10.解:原式=a(a+1)2÷a+1-1a+1=a(a+1)2×a+1a=1a+1.當a=3-1時,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.當x=2+1時,原式=22+1-1=22=2.12.解:原式=(a-b)2a·a(a-b)(a+b)=a-ba+b,當a=2-1,b=1時,原式=2-22=1-2.13.解:原式=x-2(x+1)2÷x2+xx+1-3xx+1=x-2(x+1)2·x+1x(x-2)=1x(x+1).當x=13時,原式=113×43=94.14.解:a2-4a2-4a+4-12-a÷2a2-2a=(a-2)(a+2)(a-2)2+1a-2·a(a-2)2=a+2a-2+1a-2·a(a-2)2=a+3a-2·a(a-2)2=a(a+3)2=a2+3a2,a2+3a-2=0,a2+3a=2,原式=22=1.8