课时训练05分式限时:25分钟夯实基础1.2019衡阳如果分式1x+1在实数范围内有意义,则x的取值范围是()A.x-1B.x-1C.全体实数D.x=-12.2019天津计算2aa+1+2a+1的结果是()A.2B.2a+2C.1D.4aa+13.2016桂林当x=6,y=3时,代数式xx+y+2yx+y3xyx+2y的值是()A.2B.3C.6D.94.2019贵港若分式x2-1x+1的值等于0,则x的值为()A.1B.0C.-1D.15.2017金华若ab=23,则a+bb=.6.2018永州化简:1+1x-1x2+xx2-2x+1=.7.若a=2b0,则a2-b2a2-ab的值为.8.2018玉林先化简,再求值:a-2ab-b2aa2-b2a,其中a=1+2,b=1-2.能力提升9.已知x+1x=3,则下列三个等式:x2+1x2=7;x-1x=5;2x2-6x=-2.其中正确的有()A.0个B.1个C.2个D.3个10.已知1x-1y=3,则分式2x+3xy-2yx-2xy-y的值为()A.35B.13C.53D.-5311.2018鄂州先化简,再从-3,-2,0,2中选一个合适的数作为x的值,代入求值.x2x+3x2-9x2-2x-x2x-2.12.2019桂林先化简,再求值:1y-1xx2-2xy+y22xy-1y-x,其中x=2+2,y=2.【参考答案】1.A2.A3.C解析xx+y+2yx+y3xyx+2y=x+2yx+y3xyx+2y=3xyx+y,当x=6,y=3时,原式=3636+3=6.故选C.4.D5.53解析解法1:利用比例的基本性质“两内项积等于两外项积”求解.ab=23,3a=2b.a=23b.a+bb=23b+bb=53bb=53;解法2:设参数法求解,设a=2k,则b=3k.a+bb=2k+3k3k=5k3k=53;解法3:逆用同分母分式加减法法则求解,a+bb=ab+bb=ab+1=23+1=53.6.x-1x+17.32解析a=2b0,a2-b2a2-ab=(a+b)(a-b)a(a-b)=a+ba=2b+b2b=32.8.解:原式=a2-2ab+b2aaa2-b2=(a-b)2aa(a+b)(a-b)=a-ba+b,当a=1+2,b=1-2时,原式=222=2.9.C解析x+1x=3,x2+1x2=x+1x2-2=9-2=7,正确;x-1x2=x+1x2-4=9-4=5,x-1x=5,错误;2x2-6x=-2,2x2+2=6x,又x0,两边同时除以2x,可得x+1x=3,正确.10.A解析由1x-1y=3,得y-xxy=3,即y-x=3xy.故2x+3xy-2yx-2xy-y=2(x-y)+3xy(x-y)-2xy=-6xy+3xy-3xy-2xy=-3xy-5xy=35.故选A.11.解:x2x+3x2-9x2-2x-x2x-2=x2x+3(x-3)(x+3)x(x-2)-x2x-2=x(x-3)x-2-x2x-2=x2-3x-x2x-2=-3xx-2,且x+30,x-20,x0,解得x0,x-3且x2,故当x=-2时,原式=-3(-2)(-2)-2=-32.12.解:原式=x-yxy2xy(x-y)2+1x-y=2x-y+1x-y=3x-y,当x=2+2,y=2时,原式=32+2-2=322.