第十章 專題訓(xùn)練　分式化簡求值有技巧 京改版八年級數(shù)學(xué)上冊

專題訓(xùn)練(一)分式化簡求值有技巧類型一巧代入求分式值1.已知xy=3,求x2-y2.先化簡,再求值:1+y2x2-y2·x-yx3.[2020·石景山期末]已知x2+3x=1,求代數(shù)式1x-1·x24.已知2a2-a=2,求a-2a2-5.[2020·平谷期末]已知a2-a-1=0,求代數(shù)式(1-1a+1)÷a類型二根據(jù)分式的基本性質(zhì)巧變形,求分式值6.已知a+1a=5,求a27.已知1x+1y=5,求28.已知1a+1b=1,1b+1c=2,1a+1c類型三巧設(shè)參數(shù)求分式值9.已知x4=y5=z6≠0,則210.已知x∶y=1∶2,求3x-11.已知b+ca=a+cb=類型四代入求值有陷阱12.[2020·懷柔期末改編]先化簡x-1x2+2x+1÷(13.先化簡,再求值:(1x2-2x-1x)÷x-3x214.先化簡:1-3x+2÷x2-2x+1答案1.解:x2-y2xy÷2(x由xy=3,得x=3y,把它代入x+y2x,可以求得原式=x+y2.解:1+y2x2-y2·x由x-3y=0,得x=3y,把它代入xx+y,可求出原式=xx+3.解:原式=1x-1·=x-1=x=3x∵x2+3x=1,∴原式=31+2=14.解:a-2a2-4+2a+2·由2a2-a=2,得2a2=a+2,把它代入3a2a+2,可求出原式=3a5.解:原式=(a+1a+1-1=a+1-=aa+1=aa+1=a+1由a2-a-1=0,得a2=a+1.把它代入原式,可求出原式=a+1a2=a6.[解析]若先求出a的值再代入求值,顯然以現(xiàn)在所學(xué)的知識解不出來.如果將a2a4+a2解:因為a+1a=5,所以a+則a2+2+1a2=25,所以a2+1所以a2a4+a2+17.解:由1x+1y=5,得x+yxy=所以2x-5xy+2yx+2xy8.解:方法一:由條件可知,a≠0,b≠0,c≠0,1a+1b+1b+1c+1a+1c=1所以21a+1b+1c=8,則1又因為ab+bc+acabc=ababc+bcabc+acabc=1c+1方法二:由條件可知,a≠0,b≠0,c≠0,1a+1b+1b+1c+1a+1c=1所以21a+1b+1c=8,則1又因為1a+1b=1,1b+1c=2,1所以1a=2,1b=-1,1所以a=12,b=-1,c=1將a,b,c代入abcab+bc+ac,可求出原式=-方法三:由條件可知,a≠0,b≠0,c≠0,1a+1b+1b+1c+1a+1c=1所以21a+1b+1c=8,則1根據(jù)分式的基本性質(zhì),分式的分子、分母同時除以abc,得1ababc+bcabc9.1718[解析]此題可直接令x=4k,y=5k,z=6k(k≠0),代入,得原式=2×4k-10.解:設(shè)x=k,y=2k(k≠0),故原式=3k-211.解:設(shè)b+ca=a+則b+c=ak,a+c=bk,a+b=ck.把這3個等式左右兩邊分別相加,得2(a+b+c)=(a+b+c)k.若a+b+c=0,則a+b=-c,即k=-1;若a+b+c≠0,則k=2.(a+b)(b當(dāng)k=-1時,原式=-1;當(dāng)k=2時,原式=8.12.解:原式=x-1(x+1)=x-1=1x因為分式的分母不能為0且除數(shù)不能為0,所以x不能取-1和1.例如取x=3,則原式=13+1=113.解:原式=1x(x-2)-x-因為-1<x<4,且x為整數(shù),所以x的取值可為0,1,2,3.又因為代數(shù)式(1x2-2x-1x)÷x-所以x=1.當(dāng)x=1時,原式=-1+21=-314.解:原式=x+2x=x-1