分式化簡(jiǎn)求值試題及答案

分式化簡(jiǎn)求值試題及答案

一、單項(xiàng)選擇題（每題2分，共20分）1.化簡(jiǎn)\(\frac{a^2}{a-1}-\frac{1}{a-1}\)的結(jié)果是（）A.\(a-1\)B.\(a+1\)C.\(a\)D.\(a^2-1\)2.化簡(jiǎn)\(\frac{x^2-4}{x+2}\)得（）A.\(x-2\)B.\(x+2\)C.\(x^2-4\)D.\(\frac{1}{x-2}\)3.當(dāng)\(x=2\)時(shí)，\(\frac{x^2-1}{x+1}\)的值為（）A.1B.2C.3D.44.化簡(jiǎn)\(\frac{ab}{a^2}\)的結(jié)果是（）A.\(\frac{a}\)B.\(b\)C.\(\frac{a}\)D.\(ab\)5.若分式\(\frac{x-1}{x+2}\)有意義，則\(x\)的取值范圍是（）A.\(x\neq1\)B.\(x\neq-2\)C.\(x=1\)D.\(x=-2\)6.化簡(jiǎn)\(\frac{2x}{x^2-4}-\frac{1}{x-2}\)的結(jié)果是（）A.\(\frac{1}{x+2}\)B.\(\frac{1}{x-2}\)C.\(\frac{3}{x+2}\)D.\(\frac{3}{x-2}\)7.已知\(x=3\)，\(y=2\)，則\(\frac{x^2-y^2}{x-y}\)的值為（）A.5B.1C.3D.28.化簡(jiǎn)\(\frac{a^2-9}{a^2+6a+9}\)得（）A.\(\frac{a-3}{a+3}\)B.\(\frac{a+3}{a-3}\)C.\(a-3\)D.\(a+3\)9.若\(\frac{x}{y}=\frac{2}{3}\)，則\(\frac{x+y}{y}\)的值為（）A.\(\frac{5}{3}\)B.\(\frac{2}{3}\)C.\(\frac{1}{3}\)D.\(\frac{3}{5}\)10.化簡(jiǎn)\(\frac{2}{x^2-1}+\frac{1}{x+1}\)的結(jié)果是（）A.\(\frac{3}{x+1}\)B.\(\frac{3}{x-1}\)C.\(\frac{1}{x-1}\)D.\(\frac{1}{x+1}\)二、多項(xiàng)選擇題（每題2分，共20分）1.下列分式中，是最簡(jiǎn)分式的有（）A.\(\frac{x^2}{x}\)B.\(\frac{a+b}{a^2+b^2}\)C.\(\frac{m^2-n^2}{m+n}\)D.\(\frac{y^2}{y^2+1}\)2.化簡(jiǎn)分式\(\frac{x^2-2x+1}{x^2-1}\)，正確的步驟有（）A.原式\(=\frac{(x-1)^2}{(x+1)(x-1)}\)B.當(dāng)\(x\neq1\)時(shí)，化簡(jiǎn)為\(\frac{x-1}{x+1}\)C.原式\(=\frac{x^2-2x+1}{x^2-1}\)保持不變D.當(dāng)\(x=1\)時(shí)，分式無(wú)意義3.下列分式化簡(jiǎn)正確的是（）A.\(\frac{4a^2b}{6ab^2}=\frac{2a}{3b}\)B.\(\frac{x^2-4}{x-2}=x+2\)C.\(\frac{a^2+2ab+b^2}{a+b}=a+b\)D.\(\frac{x^2}{x^2+x}=\frac{x}{x+1}\)4.計(jì)算\(\frac{1}{x-1}-\frac{1}{x+1}\)，結(jié)果可能是（）A.\(\frac{2}{x^2-1}\)B.\(\frac{2x}{x^2-1}\)C.經(jīng)過(guò)通分得到\(\frac{x+1}{(x-1)(x+1)}-\frac{x-1}{(x-1)(x+1)}\)D.\(\frac{x+1-x+1}{x^2-1}\)5.若分式\(\frac{x^2-9}{x+3}\)的值為0，則\(x\)的值可以是（）A.3B.-3C.\(x\neq-3\)D.當(dāng)\(x=3\)時(shí)滿足6.化簡(jiǎn)\(\frac{a^2-4a+4}{a^2-4}\)，說(shuō)法正確的是（）A.可化為\(\frac{(a-2)^2}{(a+2)(a-2)}\)B.當(dāng)\(a\neq2\)時(shí)，化簡(jiǎn)為\(\frac{a-2}{a+2}\)C.化簡(jiǎn)結(jié)果與\(a\)取值無(wú)關(guān)D.當(dāng)\(a=2\)時(shí)，分式無(wú)意義7.下列式子中，屬于分式的有（）A.\(\frac{1}{x}\)B.\(\frac{x}{2}\)C.\(\frac{2}{x^2+1}\)D.\(\frac{xy}{3}\)8.化簡(jiǎn)\(\frac{x^2-1}{x^2+2x+1}\)，正確的是（）A.先因式分解得\(\frac{(x+1)(x-1)}{(x+1)^2}\)B.當(dāng)\(x\neq-1\)時(shí)，化簡(jiǎn)為\(\frac{x-1}{x+1}\)C.化簡(jiǎn)后分子分母同時(shí)除以\((x+1)\)D.該分式不能再化簡(jiǎn)9.計(jì)算\(\frac{3}{x^2-4}-\frac{1}{x-2}\)，以下正確的有（）A.先通分得到\(\frac{3}{(x+2)(x-2)}-\frac{x+2}{(x+2)(x-2)}\)B.結(jié)果為\(\frac{3-x-2}{(x+2)(x-2)}=\frac{1-x}{(x+2)(x-2)}\)C.先將\(\frac{1}{x-2}\)化為\(\frac{x+2}{x^2-4}\)D.該計(jì)算無(wú)需通分10.若\(x=1\)，則下列分式的值為0的是（）A.\(\frac{x-1}{x}\)B.\(\frac{x^2-1}{x+1}\)C.\(\frac{x}{x-1}\)D.\(\frac{x^2-1}{x^2+1}\)三、判斷題（每題2分，共20分）1.分式\(\frac{1}{x}\)，當(dāng)\(x=0\)時(shí)無(wú)意義。（）2.化簡(jiǎn)\(\frac{x^2}{x}=x\)（\(x\neq0\)）。（）3.分式\(\frac{a-b}{b-a}=-1\)。（）4.若\(\frac{x}{y}=\frac{3}{4}\)，則\(\frac{x+y}{y}=\frac{7}{4}\)。（）5.化簡(jiǎn)\(\frac{x^2-16}{x-4}=x+4\)（\(x\neq4\)）。（）6.分式\(\frac{2}{x^2+1}\)無(wú)論\(x\)取何值都有意義。（）7.\(\frac{a^2+b^2}{a+b}=a+b\)。（）8.化簡(jiǎn)\(\frac{x^2-2x+1}{x-1}=x-1\)（\(x\neq1\)）。（）9.若分式\(\frac{x-3}{x^2}\)的值為0，則\(x=3\)。（）10.分式\(\frac{3}{x-2}\)與\(\frac{3x}{x^2-2x}\)是相等的分式。（）四、簡(jiǎn)答題（每題5分，共20分）1.化簡(jiǎn)\(\frac{x^2-25}{x^2+10x+25}\)并求值，其中\(zhòng)(x=3\)。答案：先因式分解，原式\(=\frac{(x+5)(x-5)}{(x+5)^2}=\frac{x-5}{x+5}\)。當(dāng)\(x=3\)時(shí)，代入得\(\frac{3-5}{3+5}=-\frac{1}{4}\)。2.化簡(jiǎn)\(\frac{3}{x^2-9}+\frac{1}{x+3}\)。答案：對(duì)\(x^2-9\)因式分解為\((x+3)(x-3)\)，則原式\(=\frac{3}{(x+3)(x-3)}+\frac{x-3}{(x+3)(x-3)}=\frac{3+x-3}{(x+3)(x-3)}=\frac{x}{(x+3)(x-3)}\)。3.已知\(\frac{x}{y}=\frac{2}{3}\)，求\(\frac{x^2-y^2}{y^2}\)的值。答案：由\(\frac{x}{y}=\frac{2}{3}\)，設(shè)\(x=2k\)，\(y=3k\)。則\(\frac{x^2-y^2}{y^2}=\frac{(2k)^2-(3k)^2}{(3k)^2}=\frac{4k^2-9k^2}{9k^2}=-\frac{5}{9}\)。4.當(dāng)\(x\)為何值時(shí)，分式\(\frac{x^2-1}{x^2-3x+2}\)無(wú)意義？答案：對(duì)分母因式分解\(x^2-3x+2=(x-1)(x-2)\)，當(dāng)\((x-1)(x-2)=0\)，即\(x=1\)或\(x=2\)時(shí)，分式無(wú)意義。五、討論題（每題5分，共20分）1.討論在化簡(jiǎn)分式\(\frac{x^2-4}{x^2-4x+4}\)時(shí)，需要注意哪些問(wèn)題？答案：先對(duì)分子分母因式分解，分子\(x^2-4=(x+2)(x-2)\)，分母\(x^2-4x+4=(x-2)^2\)，化簡(jiǎn)為\(\frac{x+2}{x-2}\)，要注意\(x\neq2\)，因?yàn)閈(x=2\)時(shí)原分式分母為0無(wú)意義。2.對(duì)于分式\(\frac{a}{a-b}\)與\(\frac{b-a}\)，在計(jì)算它們的和時(shí)，應(yīng)該怎么做？答案：先將\(\frac{b-a}\)變形為\(-\frac{a-b}\)，然后兩者相加得\(\frac{a}{a-b}-\frac{a-b}=\frac{a-b}{a-b}=1\)（\(a\neqb\)），要注意分母不能為0。3.討論如何判斷一個(gè)分式是否為最簡(jiǎn)分式？答案：看分式的分子和分母是否有公因式，若分子分母沒(méi)有公因式，則為最簡(jiǎn)分式。例如\(\frac{a+b}{a^2+b^2}\)，\(a^2+b^2\)不能分解成含\(a+b\)的因式，所以它是最簡(jiǎn)分式；若有公因式則可繼續(xù)化簡(jiǎn)。4.已知分式\(\frac{x^2+2x+1}{x^2-1}\)，當(dāng)\(x\)取不同值時(shí)，分式的值會(huì)怎樣變化？答案：先化簡(jiǎn)得\(\frac{(x+1)^2}{(x+1)(x-1)}=\frac{x+1}{x-1}=1+\frac{2}{x-1}\)（\(x\neq-1\)）。當(dāng)\(x\)增大時(shí)，\(\frac{2}{x-1}\)的值變化，分式值也變；\(x\)趨近于