分式與二次根式綜合檢測(cè)卷及答案

分式與二次根式綜合檢測(cè)卷（A卷）一、選擇題（每題3分，共30分）1.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.\(x\geq-2\)D.\(x\gt-2\)2.下列二次根式中，是最簡(jiǎn)二次根式的是（）A.\(\sqrt{12}\)B.\(\sqrt{\frac{1}{3}}\)C.\(\sqrt{30}\)D.\(\sqrt{27}\)3.計(jì)算\(\frac{1}{a-1}-\frac{1}{a^2-1}\)的結(jié)果是（）A.\(\frac{a}{a^2-1}\)B.\(\frac{a}{a-1}\)C.\(\frac{1}{a+1}\)D.\(\frac{a+1}{a}\)4.若\(\sqrt{(x-3)^2}=3-x\)，則\(x\)的取值范圍是（）A.\(x\lt3\)B.\(x\leq3\)C.\(x\gt3\)D.\(x\geq3\)5.化簡(jiǎn)\(\frac{x^2-4}{x^2-4x+4}\div\frac{x+2}{x-2}\)的結(jié)果是（）A.\(0\)B.\(1\)C.\(x+2\)D.\(x-2\)6.已知\(a=\sqrt{3}+1\)，\(b=\sqrt{3}-1\)，則\(a^2-b^2\)的值為（）A.\(4\sqrt{3}\)B.\(4\)C.\(2\sqrt{3}\)D.\(2\)7.若分式\(\frac{x^2-9}{x+3}\)的值為\(0\)，則\(x\)的值是（）A.\(3\)B.\(-3\)C.\(\pm3\)D.\(0\)8.計(jì)算\((\sqrt{5}+2)^{2023}(\sqrt{5}-2)^{2022}\)的結(jié)果是（）A.\(\sqrt{5}+2\)B.\(\sqrt{5}-2\)C.\(1\)D.\(-1\)9.若\(m\lt0\)，化簡(jiǎn)\(\frac{\vertm\vert}{m}+\frac{m+\vertm\vert}{m}\)的結(jié)果為（）A.\(0\)B.\(-1\)C.\(-2\)D.\(2\)10.已知\(x=\frac{1}{\sqrt{5}-2}\)，\(y=\frac{1}{\sqrt{5}+2}\)，則\(x^2+xy+y^2\)的值為（）A.\(18\)B.\(17\)C.\(16\)D.\(15\)二、填空題（每題3分，共18分）11.當(dāng)\(x=\)______時(shí)，分式\(\frac{1}{x-2}\)與\(\frac{3}{x}\)的值相等。12.計(jì)算：\(\sqrt{8}\times\sqrt{\frac{1}{2}}=\)______。13.化簡(jiǎn)：\(\frac{2}{x^2-4}-\frac{1}{2x-4}=\)______。14.已知\(a\)，\(b\)為實(shí)數(shù)，且\(\sqrt{a-3}+(b+2)^2=0\)，則\(b^a=\)______。15.若\(\sqrt{20n}\)是整數(shù)，則正整數(shù)\(n\)的最小值為______。16.已知\(x=\sqrt{3}-1\)，則\(x^2+2x+3\)的值為______。三、解答題（共52分）17.（8分）計(jì)算：（1）\(\sqrt{27}-\sqrt{12}+\sqrt{\frac{1}{3}}\)（2）\((\frac{1}{x-1}-\frac{1}{x+1})\div\frac{2}{x^2-1}\)18.（8分）化簡(jiǎn)求值：\(\frac{x^2-1}{x^2+2x+1}\div\frac{x-1}{x+1}-\frac{x}{x+1}\)，其中\(zhòng)(x=\sqrt{2}-1\)。19.（8分）解方程：\(\frac{2}{x-3}=\frac{3}{x}\)20.（8分）已知\(a=\frac{1}{\sqrt{3}+\sqrt{2}}\)，\(b=\frac{1}{\sqrt{3}-\sqrt{2}}\)，求\(a^2-ab+b^2\)的值。21.（10分）某商場(chǎng)銷售某種商品，第一個(gè)月將此商品的進(jìn)價(jià)提高百分之二十作為售價(jià)，共獲利6000元，第二個(gè)月商場(chǎng)搞促銷活動(dòng)，將商品的進(jìn)價(jià)提高百分之十作為售價(jià)，第二個(gè)月比第一個(gè)月銷售量增加了100件，并且第二個(gè)月比第一個(gè)月多獲利2000元，問此商品的進(jìn)價(jià)每件是多少元？商場(chǎng)第二個(gè)月共銷售商品多少件？22.（10分）觀察下列各式：\(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}=1+\frac{1}{1\times2}=1+1-\frac{1}{2}=\frac{3}{2}\)\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}=1+\frac{1}{2\times3}=1+\frac{1}{2}-\frac{1}{3}=\frac{7}{6}\)\(\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}=1+\frac{1}{3\times4}=1+\frac{1}{3}-\frac{1}{4}=\frac{13}{12}\)……（1）根據(jù)上述規(guī)律，寫出第\(n\)個(gè)等式，并證明你的結(jié)論。（2）利用上述規(guī)律計(jì)算：\(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+\cdots+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}\)答案一、選擇題1.B2.C3.A4.B5.B6.A7.A8.A9.C10.B二、填空題11.\(3\)12.\(2\)13.\(-\frac{1}{2(x+2)}\)14.\(-8\)15.\(5\)16.\(5\)三、解答題17.（1）原式\(=3\sqrt{3}-2\sqrt{3}+\frac{\sqrt{3}}{3}=\frac{4\sqrt{3}}{3}\)（2）原式\(=\left[\frac{x+1-(x-1)}{(x-1)(x+1)}\right]\div\frac{2}{(x+1)(x-1)}=\frac{2}{(x-1)(x+1)}\times\frac{(x+1)(x-1)}{2}=1\)18.原式\(=\frac{(x+1)(x-1)}{(x+1)^2}\times\frac{x+1}{x-1}-\frac{x}{x+1}=1-\frac{x}{x+1}=\frac{1}{x+1}\)當(dāng)\(x=\sqrt{2}-1\)時(shí)，原式\(=\frac{1}{\sqrt{2}-1+1}=\frac{\sqrt{2}}{2}\)19.方程兩邊同乘\(x(x-3)\)得：\(2x=3(x-3)\)\(2x=3x-9\)\(3x-2x=9\)\(x=9\)經(jīng)檢驗(yàn)，\(x=9\)是原方程的解。20.\(a=\frac{1}{\sqrt{3}+\sqrt{2}}=\sqrt{3}-\sqrt{2}\)，\(b=\frac{1}{\sqrt{3}-\sqrt{2}}=\sqrt{3}+\sqrt{2}\)\(a+b=2\sqrt{3}\)，\(ab=1\)\(a^2-ab+b^2=(a+b)^2-3ab=(2\sqrt{3})^2-3\times1=12-3=9\)21.設(shè)此商品進(jìn)價(jià)為每件\(x\)元。\(\frac{6000}{0.2x}+100=\frac{6000+2000}{0.1x}\)\(\frac{30000}{x}+100=\frac{80000}{x}\)\(80000-30000=100x\)\(100x=50000\)\(x=500\)第二個(gè)月銷售量為\(\frac{6000+2000}{0.1\times500}=160\)（件）答：此商品進(jìn)價(jià)每件\(500\)元，商場(chǎng)第二個(gè)月共銷售商品\(160\)件。22.（1）第\(n\)個(gè)等式為：\(\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}=1+\frac{1}{n(n+1)}\)證明：\[\begin{align}\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}&=\sqrt{\frac{n^2(n+1)^2+(n+1)^2+n^2}{n^2(n+1)^2}}\\&=\sqrt{\frac{[n(n+1)]^2+2n^2+2n+1}{n^2(n+1)^2}}\\&=\sqrt{\frac{[n(n+1)]^2+2n(n+1)+1}{n^2(n+1)^2}}\\&=\sqrt{\frac{(n^2+n+1)^2}{n^2(n+1)^2}}\\&=\frac{n^2+n+1}{n(n+1)}\\&=\frac{n(n+1)+1}{n(n+1)}\\&=1+\frac{1}{n(n+1)}\end{align}\]（2）原式\(=(1+1-\frac{1}{2})+(1+\frac{1}{2}-\frac{1}{3})+(1+\frac{1}{3}-\frac{1}{4})+\cdots+(1+\frac{1}{99}-\frac{1}{100})\)\(=99+(1-\frac{1}{100})\)\(=99+\frac{99}{100}\)\(=99\frac{99}{100}\)分式與二次根式綜合檢測(cè)卷（B卷）一、選擇題（每題3分，共30分）1.使二次根式\(\sqrt{x-3}\)有意義的\(x\)的取值范圍是（）A.\(x\gt3\)B.\(x\geq3\)C.\(x\lt3\)D.\(x\leq3\)2.下列分式中，無論\(x\)取何值，分式都有意義的是（）A.\(\frac{1}{x^2}\)B.\(\frac{x}{x^2-1}\)C.\(\frac{3x}{x^2+1}\)D.\(\frac{x+1}{x}\)3.計(jì)算\(\frac{a}{a-b}-\frac{a-b}\)的結(jié)果是（）A.\(1\)B.\(-1\)C.\(\frac{a+b}{a-b}\)D.\(\frac{a-b}{a+b}\)4.若\(\sqrt{12n}\)是正整數(shù)，則\(n\)的最小值為（）A.\(1\)B.\(2\)C.\(3\)D.\(4\)5.化簡(jiǎn)\(\frac{x^2-1}{x}\div\frac{x+1}{x^2}\)的結(jié)果是（）A.\(x^2-x\)B.\(x^2+x\)C.\(x-1\)D.\(x+1\)6.已知\(x=\sqrt{2}-1\)，\(y=\sqrt{2}+1\)，則\(\frac{1}{x}+\frac{1}{y}\)的值為（）A.\(2\sqrt{2}\)B.\(-2\sqrt{2}\)C.\(2\)D.\(-2\)7.若分式\(\frac{\vertx\vert-3}{x+3}\)的值為\(0\)，則\(x\)的值為（）A.\(3\)B.\(-3\)C.\(\pm3\)D.無解8.計(jì)算\((\sqrt{3}-2)^{2022}(\sqrt{3}+2)^{2023}\)的結(jié)果是（）A.\(\sqrt{3}-2\)B.\(\sqrt{3}+2\)C.\(1\)D.\(-1\)9.化簡(jiǎn)\(\frac{m^2-4m+4}{m^2-4}\div\frac{m-2}{m^2+2m}\)的結(jié)果為（）A.\(m\)B.\(m+2\)C.\(m-2\)D.\(\frac{m}{m-2}\