1、2021-2022年四川省中考数学真题分类专题3分式、二次根式一选择题(共11小题)1(2022眉山)化简4a+2+a2的结果是()A1Ba2a+2Ca2a2-4Daa+22(2022凉山州)分式13+x有意义的条件是()Ax3Bx3Cx3Dx03(2022德阳)下列计算正确的是()A(ab)2a2b2B(-1)2=1Caa1a=aD(-12ab2)3=-16a3b64(2022自贡)下列运算正确的是()A(1)22B(3+2)(3-2)1Ca6a3a2D(-12022)005(2022南充)已知ab0,且a2+b23ab,则(1a+1b)2(1a2-1b2)的值是()A5B-5C55D-55
2、6(2021巴中)下列各式的值最小的是()A20B|2|C21D(2)7(2021南充)下列运算正确的是()A3b4a2a9b2=b6B13ab2b23a=b32C12a+1a=23aD1a-1-1a+1=2a2-18(2021雅安)若分式|x|-1x-1的值等于0,则x的值为()A1B0C1D19(2021内江)函数y=2-x+1x+1中,自变量x的取值范围是()Ax2Bx2且x1Cx2Dx2且x110(2021绵阳)计算1812的结果是()A6B62C63D6611(2021达州)下列计算正确的是()A2+3=5B(-3)2=3Caa11(a0)D(3a2b2)26a4b4二填空题(共7小
3、题)12(2022南充)比较大小:22 30(选填,)13(2022自贡)化简:a-3a2+4a+4a2-4a-3+2a+2= 14(2021南充)若n+mn-m=3,则m2n2+n2m2= 15(2021乐山)(2021)0 16(2022遂宁)实数a、b在数轴上的位置如图所示,化简|a+1|-(b-1)2+(a-b)2= 17(2021资阳)若x2+x10,则3x-3x= 18(2021自贡)化简:2a-2-8a2-4= 三解答题(共10小题)19(2022广元)先化简,再求值:2x2+x(1-x-1x2-1),其中x是不等式组2(x-1)x+15x+32x的整数解20(2022乐山)先化
4、简,再求值:(1-1x+1)xx2+2x+1,其中x=221(2022达州)化简求值:a-1a2-2a+1(a2+aa2-1+1a-1),其中a=3-122(2022凉山州)先化简,再求值:(m+2+52-m)2m-43-m,其中m为满足1m4的整数23(2022泸州)化简:(m2-3m+1m+1)m2-1m24(2022遂宁)先化简,再求值:(1-2a+1)2a2-2a+1a+1,其中a425(2021绵阳)(1)计算:2cos45+|2-3|20210-33;(2)先化简,再求值:2x-y-xx+y-2xyx2-y2,其中x1.12,y0.6826(2021巴中)(1)计算:2sin60+
5、|3-2|(12)1+6-22;(2)解不等式组2x+5-123x-3x+1213,并把解集在数轴上表示出来(3)先化简,再求值:a2+8a+16a2+3a(1+1a+3),请从4,3,0,1中选一个合适的数作为a的值代入求值27(2021雅安)(1)计算:(12)2+(3.14)0+|3-12|4sin60(2)先化简,再求值:(1x-1-x+1)x-2x2-1,其中x=2-128(2021广元)先化简,再求值:(1x-y+1x+y)1x2+xy其中x=2,y12021-2022年四川省中考数学真题分类专题3分式、二次根式参考答案与试题解析一选择题(共11小题)1【解答】解:4a+2+a-2
6、=4a+2+a2-4a+2 =a2a+2故选:B2【解答】解:由题意得:3+x0,x3,故选:B3【解答】解:A(ab)2a22ab+b2,故A选项错误,不符合题意;B.(-1)2=12=1,故B选项正确,符合题意;Caa1a=11a=1a,故C选项错误,不符合题意;D(-12ab2)3=-18a3b6,故D选项错误,不符合题意故选:B4【解答】解:A、原式1,故该选项不符合题意;B、原式(3)2(2)2321,故该选项符合题意;C、原式a3,故该选项不符合题意;D、原式1,故该选项不符合题意;故选:B5【解答】解:(1a+1b)2(1a2-1b2)=(a+b)2a2b2b2-a2a2b2 =
7、(a+b)2a2b2a2b2(b+a)(b-a) =-a+ba-b,a2+b23ab,(a+b)25ab,(ab)2ab,ab0,a+b=5ab,ab=ab,-a+ba-b=-5abab=-5abab=-5,故选:B6【解答】解:201,|2|2,21=12,(2)2,1212,最小的是21故选:C7【解答】解:3b4a2a9b2=16b,故选项A错误;13ab2b23a=13ab3a2b2=12b3,故选项B错误;12a+1a=12a+22a=32a,故选项C错误;1a-1-1a+1=a+1-(a-1)(a+1)(a-1)=a+1-a+1(a+1)(a-1)=2a2-1,故选项D正确;故选:
8、D8【解答】解:由题意得:|x|10,且x10,解得:x1,故选:A9【解答】解:由题意得:2x0,x+10,解得:x2且x1,故选:B10【解答】解:1812=1812 =9243 66,故选:D11【解答】解:A.2+3无法合并,故此选项错误;B.(-3)2=3,故此选项错误;Caa1=aa=1(a0),故此选项正确;D(3a2b2)29a4b4,故此选项错误;故选:C二填空题(共7小题)12【解答】解:22=14,301,2230,故答案为:13【解答】解:a-3a2+4a+4a2-4a-3+2a+2=a-3(a+2)2(a+2)(a-2)a-3+2a+2 =a-2a+2+2a+2 =a
9、a+2,故答案为:aa+214【解答】解:n+mn-m=3,n2m,m2n2+n2m2=m2(2m)2+(2m)2m2=14+4=174,故答案为:17415【解答】解:(2021)01故答案为:116【解答】解:由数轴可得,1a0,1b2,a+10,b10,ab0,|a+1|-(b-1)2+(a-b)2a+1(b1)+(ba)a+1b+1+ba2,故答案为:217【解答】解:3x-3x=3(x-1x),x2+x10(x0),x+1-1x=0,x-1x=-1,当x-1x=-1时,原式3(1)3,故答案为:318【解答】解:2a-2-8a2-4=2a-2-8(a+2)(a-2) =2(a+2)(
10、a+2)(a-2)-8(a+2)(a-2) =2a-4(a-2)(a+2) =2(a-2)(a-2)(a+2) =2a+2故答案为:2a+2三解答题(共10小题)19【解答】解:原式=2x(x+1)x2-1-x+1(x+1)(x-1)=2x(x+1)(x+1)(x-1)x(x-1) =2x2,解第一个不等式得:x3,解第二个不等式得:x1,不等式组的解集为:1x3,x为整数,x的值为1,0,1,2,x0,x+10,(x+1)(x1)0,x(x1)0,x只能取2,当x2时,原式=222=1220【解答】解:(1-1x+1)xx2+2x+1=x+1-1x+1(x+1)2x =xx+1(x+1)2x
11、 x+1,当x=2时,原式=2+121【解答】解:原式=a-1(a-1)2a(a+1)(a-1)(a+1)+a+1(a-1)(a+1)=1a-1(a+1)2(a-1)(a+1) =1a-1a+1a-1 =1a-1a-1a+1 =1a+1,把a=3-1代入1a+1=13-1+1=3322【解答】解:(m+2+52-m)2m-43-m=m2-4-5m-22(m-2)3-m =m2-9m-22(m-2)3-m =(m+3)(m-3)m-22(m-2)3-m 2(m+3)2m6,m2,m3,当m1时,原式21626823【解答】解:原式=m2-3m+1+mmm2-1m=m2-2m+1mmm2-1 =(
12、m-1)2mm(m+1)(m-1) =m-1m+124【解答】解:原式=(a+1a+1-2a+1)2a+1(a-1)2=(a-1a+1)2a+1(a-1)2 =1a+1当a4时,原式=14+1=1525【解答】解:(1)原式222+3-2-1-3=2+3-2-1-3 1,(2)原式=2(x+y)x2-y2-x(x-y)x2-y2-2xyx2-y2=2(x+y)-x(x+y)x2-y2 =(2-x)(x+y)(x+y)(x-y) =2-xx-y,当x1.12,y0.68时:2-xx-y=2-1.121.12-0.68=226【解答】解:(1)2sin60+|3-2|(12)1+6-22232+2
13、-3-2+3-1=3+2-3-2+3-1=3-1;(2)2x+5-123x-3x+1213,解不等式,得x3,解不等式,得x1,原不等式组的解集是3x1,解集在数轴上表示如下:;(3)a2+8a+16a2+3a(1+1a+3)=(a+4)2a(a+3)a+3+1a+3 =(a+4)2a(a+3)a+3a+4 =a+4a,a(a+3)0,a+40,a4,3,0,a1,当a1时,原式=1+41=527【解答】解:原式4+1+12-34325+23-3232(2)原式1x-1-(x-1)2x-1(x-1)(x+1)x-2=1-x2+2x-1x-1(x-1)(x+1)x-2 =-x(x-2)x-1(x-1)(x+1)x-2 x(x+1)x2x,当x=2-1时,x+1=2,原式=-2(2-1)2+228【解答】解:(1x-y+1x+y)1x2+xy=x+y+x-y(x+y)(x-y)x(x+y)=2xx-yx=2x2x-y,当x=2,y1时,原式=2(2)22-1=42+4