中考數(shù)學(xué)考點(diǎn)集訓(xùn)分類訓(xùn)練3 分式(含答案)

分類訓(xùn)練3分式命題點(diǎn)1分式有意義的條件1(2022黃岡)若分式2x-1有意義,則x的取值范圍是命題點(diǎn)2分式值為0的條件2(2022廣西北部灣經(jīng)濟(jì)區(qū))當(dāng)x=時(shí),分式2xx命題點(diǎn)3分式的化簡(jiǎn)3(2022山西)化簡(jiǎn)1a-3-6a2-9A.1a+3 B.a-3 C.a+34(2022威海)試卷上一個(gè)正確的式子(1a+b+1a-b)÷★=2aA.aa-b B.a-b5(2022蘇州)化簡(jiǎn)x2x-2-26(2022自貢)化簡(jiǎn):a-3a2+4a+4·7(2022河南)化簡(jiǎn):x2-1x÷(1-8(2022天門(mén))化簡(jiǎn):(m2-9m2-6m9(2022泰安)化簡(jiǎn):(a-2-4a-2)÷命題點(diǎn)4分式的化簡(jiǎn)求值角度1已知字母的值10(2022株洲)先化簡(jiǎn),再求值:(1+1x+1)·x+1x2+411(2022達(dá)州)化簡(jiǎn)求值:a-1a2-2a+1÷(a2+aa角度2結(jié)合實(shí)數(shù)的運(yùn)算12(2022濱州)先化簡(jiǎn),再求值:(a+1-3a-1)÷a2+4a+4a-1,其中a=tan45°+(13(2022荊州)先化簡(jiǎn),再求值:(aa2-b2-1a+b)÷ba2-2ab+b2,其中角度3結(jié)合方程14(2022南充)已知a>b>0,且a2+b2=3ab,則(1a+1b)2÷(1a2-1b2A.5 B.-5C.55 D.-15(2022宜昌)求代數(shù)式3x+2yx2-y2+角度4結(jié)合不等式組16(2022廣元)先化簡(jiǎn),再求值:2x2+x÷(1-x-1x2角度5自選值代入17(2022邵陽(yáng))先化簡(jiǎn),再?gòu)?1,0,1,3中選擇一個(gè)合適的x值代入求值.(1x+1+1x2分類訓(xùn)練3分式1.x≠12.03.A【解析】原式=a+3(a-3)(a+3)-4.A【解析】2a+b≠0,(1a+b+1a-b)÷25.x【解析】原式=x2-2x6.aa+2【解析】原式=a-3(a+2)2·(a+2)(7.【參考答案】原式=(x+1)(=x+1.8.【參考答案】原式=[(m+3)(m-3)=(m+3m-3-=mm-3=1m9.【參考答案】原式=(a-2=a2-4=a(a=a(a+2)=a2+2a.10.【參考答案】原式=(x+1x+1+=x+2x=1x∵x=4,∴原式=14+2=111.【參考答案】原式=a-1(=a-1(=1a當(dāng)a=3-1時(shí),原式=13-1+1=12.【參考答案】原式=(a2-1a-1-=a2-4=(a+2)(=a-2∵a=tan45°+(12)-1-π0=1+2-1=∴原式=2-22+2=013.【參考答案】原式=[a(a+b=a(a+b)(a=a2-=b=a-∵a=(13)-1=3,b=(-2022)0=∴原式=3-13+1=114.B【解析】原式=(a+b)2a2b2÷(-a2-b2a2b2)=-(a+b)2a2b2·a2b2(a+b)(a-b)=-a+ba-b.∵a2+b2=3ab,∴a2+b2+2ab=3ab+2ab,a2+b2-215.【參考答案】原式=3x+2yx2-=2x當(dāng)x=2+y時(shí),原式=22=116.【參考答案】原式=2x(=2x(=2x解2(x-1)<x+1,得x<3,解5x+3≥2x,得x≥-1,故不等式組的解集為-1≤x<3,其整數(shù)解為-1,0,1,2.∵當(dāng)x=-1,0