北師大版八年級下冊《第5章 分式與分式方程》單元檢測卷A（一）

1、北師大版八年級下冊第5章 分式與分式方程2014年單元檢測卷a（一）一、選擇題（每小題4分，共48分）1下列各式x23，1y+2，a21a1，3m中，分式有（）a1個(gè)b2個(gè)c3個(gè)d4個(gè)2分式x21x+1的值為零，則x的值為（）a-1b0c1d13把分式ab+c的a、b、c的值都擴(kuò)大為原來的3倍，則分式的值（）a不變b變?yōu)樵瓉淼?倍c變?yōu)樵瓉淼?3d變?yōu)樵瓉淼?64下列計(jì)算錯(cuò)誤的是（）a0.2a+b0.7ab2a+b7abbx3y2x2y3xycabba1d1c+2c3c5化簡2xx29+13x的結(jié)果是（）a1x3b1x+3c13xd3x+3x296化簡（x-2x1x）（1-1x）的結(jié)果是（）a

2、1xbx-1cx1xdxx17為了幫助遭受自然災(zāi)害的地區(qū)重建家園，某學(xué)校號召同學(xué)們自愿捐款已知第一次捐款總額為4800元，第二次捐款總額為5000元，第二次捐款人數(shù)比第一次多20人，而且兩次人均捐款額恰好相等，如果設(shè)第一次捐款人數(shù)是x人，那么x滿足的方程是（）a4800x5000x20b4800x=5000x+20c4800x205000xd4800x+205000x8分式方程1x1+2xx+1=2的解是（）a1b-1c3d無解9對于非零的實(shí)數(shù)a、b，規(guī)定ab=1b-1a若2（2x-1）=1，則x=（）a56b54c32d-1610若方程xx3-2=kx3有增根，則k=（）a-2b0c1d31

3、1關(guān)于x的分式方程mx+11的解是負(fù)數(shù)，則m的取值范圍是（）am-1bm-1且m0cm-1dm-1且m012如圖，設(shè)k=甲圖中陰影部分面積乙圖中陰影部分面積（ab0），則有（）ak2b1k2c12k1d0k12二、填空題（每小題4分，共24分）13要使分式2xx3有意義，則x須滿足的條件為14計(jì)算：(1+4a24)aa2=15杭州到北京的鐵路長1487千米火車的原平均速度為x千米/時(shí)，提速后平均速度增加了70千米/時(shí)，由杭州到北京的行駛時(shí)間縮短了3小時(shí)，則可列方程為16已知1a1b12，則abab的值是17若關(guān)于x的方程axx2=4x2+1無解，則a的值是18已知三個(gè)數(shù)x，y，z滿足xyx+y

4、2，yzy+z43，zxz+x43，則xyzxy+yz+zx的值為三、解答題（19題6分，20題8分，共14分）19計(jì)算：（1）2aa291a3；（2）(1a1b)a2b2ab20解分式方程：（1）23x1136x2；（2）3x1x+3x10四、解答題（每小題10分，共40分）21先化簡，再求值：aba（a-2abb2a），其中a、b滿足式子|a-2|+（b-3）2=022我市新城區(qū)環(huán)形路的拓寬改造工程項(xiàng)目，經(jīng)投標(biāo)決定由甲、乙兩個(gè)工程隊(duì)共同完成這一工程項(xiàng)目已知乙隊(duì)單獨(dú)完成這項(xiàng)工程所需天數(shù)是甲隊(duì)單獨(dú)完成這項(xiàng)工程所需天數(shù)的2倍；該工程如果由甲隊(duì)先做6天，剩下的工程再由甲、乙兩隊(duì)合作16天可以完成求

5、甲、乙兩隊(duì)單獨(dú)完成這項(xiàng)工程各需要多少天？23a玉米試驗(yàn)田是邊長為a米的正方形減去一個(gè)邊長為1米的正方形蓄水池后余下部分，b玉米試驗(yàn)田是邊長為（a-1）米的正方形，兩塊試驗(yàn)田的玉米都收獲了500千克（1）哪種玉米的單位面積產(chǎn)量高？（2）高的單位面積產(chǎn)量是低的單位面積產(chǎn)量的多少倍？24興發(fā)服裝店老板用4500元購進(jìn)一批某款t恤衫，由于深受顧客喜愛，很快售完，老板又用4950元購進(jìn)第二批該款式t恤衫，所購數(shù)量與第一批相同，但每件進(jìn)價(jià)比第一批多了9元（1）第一批該款式t恤衫每件進(jìn)價(jià)是多少元？（2）老板以每件120元的價(jià)格銷售該款式t恤衫，當(dāng)?shù)诙鷗恤衫售出45時(shí)，出現(xiàn)了滯銷，于是決定降價(jià)促銷，若要使第

6、二批的銷售利潤不低于650元，剩余的t恤衫每件售價(jià)至少要多少元？（利潤=售價(jià)-進(jìn)價(jià)）五、解答題（每小題12分，共24分）25先觀察下列等式，然后用你發(fā)現(xiàn)的規(guī)律解答下列問題11211212312131341314（1）計(jì)算112+123+134+145+156=；（2）探究112+123+134+1n(n+1)=；（用含有n的式子表示）（3）若113+135+157+1(2n1)(2n+1)的值為1735，求n的值26閱讀下面材料，并解答問題材料：將分式x4x2+3x2+1拆分成一個(gè)整式與一個(gè)分式（分子為整數(shù)）的和的形式解：由分母為-x2+1，可設(shè)-x4-x2+3=（-x2+1）（x2+a）+b

7、則-x4-x2+3=（-x2+1）（x2+a）+b=-x4-ax2+x2+a+b=-x4-（a-1）x2+（a+b）對應(yīng)任意x，上述等式均成立，a11a+b3，a=2，b=1x4x2+3x2+1(x2+1)(x2+2)+1x2+1=(x2+1)(x2+2)x2+1+1x2+1=x2+2+1x2+1這樣，分式x4x2+3x2+1被拆分成了一個(gè)整式x2+2與一個(gè)分式1x2+1的和解答：（1）將分式x46x2+8x2+1拆分成一個(gè)整式與一個(gè)分式（分子為整數(shù)）的和的形式（2）當(dāng)x（-1，1），試說明x46x2+8x2+1的最小值為8 tgz7r4i30ka1dkaghn3xtkknbycudxqa7f

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