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1、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+y·3xyx+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=
2、 . 6.[2018·永州]化简:1+1x-1÷x2+xx2-2x+1= . 7.若a=2b≠0,则a2-b2a2-ab的值为 . 8.[2018·玉林]先化简,再求值:a-2ab-b2a÷a2-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:①x2+1x2=7;②x-1x=5;③2x2-6x=-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
3、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+3·x2-9x2-2x-x2x-2.12.[2019·桂林]先化简,再求值:1y-1x÷x2-2xy+y22xy-1y-x,其中x=2+2,y=2.【参考答案】1.A 2.A3.C [解析]xx+y+2yx+y·3xyx+2y=x+2yx+y·3xyx+2y=3xyx+y,当x=6,y=3时,原式=3×6×36+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=23
4、b+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=2b≠0,∴a2-b2a2-ab=(a+b)(a-b)a(a-b)=a+ba=2b+b2b=32.8.解:原式=a2-2ab+b2a·aa2-b2=(a-b)2a·a(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+
5、1x2-2=9-2=7,①正确;∵x-1x2=x+1x2-4=9-4=5,∴x-1x=±5,②错误;∵2x2-6x=-2,∴2x2+2=6x,又∵x≠0,∴两边同时除以2x,可得x+1x=3,③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+3·x2-9x2-2x-x2x-2=x2x+3·(x-3)(x+3)x(x-2)-x2x-2=x(x-3)x-
6、2-x2x-2=x2-3x-x2x-2=-3xx-2,且x+3≠0,x-2≠0,x≠0,解得x≠0,x≠-3且x≠2,故当x=-2时,原式=-3×(-2)(-2)-2=-32.12.解:原式=x-yxy·2xy(x-y)2+1x-y=2x-y+1x-y=3x-y,当x=2+2,y=2时,原式=32+2-2=322.
