【文档说明】浙教版七年级数学下册第五章分式5.4分式的加减一练习(含答案).doc,共(5)页,56.000 KB,由MTyang资料小铺上传
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5.4分式的加减(一)A组1.计算aa+1+1a+1的结果是(A)A.1B.aC.a+1D.1a+12.下列等式成立的是(C)A.1a+2b=3a+bB.22a+b=1a+bC.abab-b2=aa-bD.a-a+b=-aa+b3
.计算:(1)x-yx+x+yx=__2__.(2)xx-1-1x-1=__1__.(3)x2x+1-1x+1=__x-1__.4.化简2xx+1+1-xx+1的结果是__1__.5.计算:(1)x2x-2+4x2-x+4x-2.【解】原式=x2-4x+4x-2=(x-2)2x-2=x-2
.(2)-2a+3ba-b+2b+aa-b-3b-ab-a.【解】原式=2a+3bb-a-a+2bb-a-3b-ab-a=(2a+3b)-(a+2b)-(3b-a)b-a=2a-2bb-a=2(a-b)b-a=-2.(3)-xx-2-1+x2-x+x2-5x
-2.【解】原式=-xx-2+x+1x-2+x2-5x-2=-x+(x+1)+(x2-5)x-2=x2-4x-2=x+2.(4)x-1x+2÷x2-2x+1x2-4-1x-1.【解】原式=x-1x+2·(x+2)(x-2)(x-1)2-1x-1
=x-2x-1-1x-1=x-3x-1.6.先化简,再求值:x+2x-2-x-1x2-4÷1x+2,其中x=-1.【解】原式=x+2x-2-x-1(x+2)(x-2)·(x+2)=x+2x-2-x-1x-2=x+2-x+1x-2=3x-2.当x=-1时,原式=3-1
-2=-1.B组7.化简2a-1-a+1a2-2a+1÷a+1a-1的结果是(B)A.1a+1B.1a-1C.2a+1D.3a-1【解】原式=2a-1-a+1(a-1)2·a-1a+1=2a-1-1a-1=1a-1.8.化简1a+1b÷1a2-1b2+1a-b的结果是
(B)A.ab+1a-bB.ab-1b-aC.abb-aD.ab+1b-a【解】原式=b+aab÷b2-a2a2b2+1a-b=b+aab·a2b2(b+a)(b-a)+1a-b=ab-1b-a.9.计算:(1)x+2yx2-y2+yy2-x2-2xx2-y2.【解】原式
=x+2yx2-y2-yx2-y2-2xx2-y2=x+2y-y-2xx2-y2=-x+y(x+y)(x-y)=-1x+y.(2)2a-3a+1-(a-2)(a-1)a2-1.【解】原式=2a-3a+1-a-2a+1=2a-3-a+2a+1=a-1a+1.(3)a2a-2+42-
a·1a2+2a.【解】原式=a2a-2-4a-2·1a(a+2)=(a+2)(a-2)a-2·1a(a+2)=1a.10.先化简,再求值:1x-y+1x+y÷2x-yx2-y2,其中x,y满足|x-2|+(2x-y-3)2=0.【解】原式=x+y+
x-y(x-y)(x+y)·(x+y)(x-y)2x-y=2x2x-y.∵|x-2|+(2x-y-3)2=0,∴x-2=0,2x-y-3=0,解得x=2,y=1.∴原式=2×22×2-1
=43.11.从甲地到乙地有两条路,每条路都是6km,其中第一条路是平路,第二条路有3km的上坡路、3km的下坡路.小丽在上坡路上的骑车速度为v(km/h),在平路上的骑车速度为2v(km/h),在下坡路上的骑车
速度为3v(km/h).(1)当走第二条路时,她从甲地到乙地需要多少时间?(2)她走哪条路花费的时间少?少多长时间?【解】(1)当走第二条路时,从甲地到乙地需要的时间为3v+33v=3v+1v=4v(h).(2)当走第一条路时,从
甲地到乙地需要的时间为62v=3v(h).∵4v-3v=1v(h),∴小丽走第一条路花费的时间少,少1vh.12.先化简a2+4aa-2-42-a·a-2a2-4,再从1,2,3中选取一个
适当的数代入求值.【解】a2+4aa-2-42-a·a-2a2-4=a2+4a+4a-2·a-2a2-4=(a+2)2a-2·a-2(a+2)(a-2)=a+2a-2.∵a-2≠0,a+2≠0,
∴a≠±2,∴a可取1或3.当a=1时,原式=-3;当a=3时,原式=5.13.已知abc=1,求aab+a+1+bbc+b+1+cac+c+1的值.【解】原式=ac(ab+a+1)c+bbc+b+abc+cac+c+1=acabc+ac+c+bb(c+1+ac)+cac+c+1=ac1
+ac+c+1c+1+ac+cac+c+1=ac+c+1ac+c+1=1.数学乐园14.当x分别取-2018,-2017,-2016,…,-2,-1,0,1,12,13,…,12016,12017,12018时,计算分式x2-1x2
+1的值,再将所得结果相加,其和为多少?【解】设a为负整数.∵当x=a时,x2-1x2+1=a2-1a2+1;当x=-1a时,x2-1x2+1=-1a2-11a2+1=1-a2a2+1,a2-1
a2+1+1-a2a2+1=0,∴当x的值互为负倒数时,两分式的和为0,∴所得结果的和=02-102+1=-1.