八年级分式试卷及答案

八年级分式试卷及答案一、选择题（每题3分，共30分）1.下列分式中，分母为多项式的是（）A.\(\frac{1}{2}\)B.\(\frac{1}{x+1}\)C.\(\frac{1}{x^2-1}\)D.\(\frac{1}{x}\)2.计算\(\frac{a}{b}\div\frac{c}{d}\)的结果是（）A.\(\frac{ac}{bd}\)B.\(\frac{ad}{bc}\)C.\(\frac{ad}{cb}\)D.\(\frac{ac}{bd}\)3.如果\(\frac{a}{b}=\frac{c}{d}\)，那么\(ad=bc\)（）A.正确B.错误4.将分式\(\frac{2x}{x-1}\)化简为最简分式，结果是（）A.\(2\)B.\(2+\frac{2}{x-1}\)C.\(\frac{2x}{x-1}\)D.\(\frac{2}{1-x}\)5.计算\(\frac{1}{x-1}+\frac{1}{x+1}\)的结果是（）A.\(\frac{2}{x^2-1}\)B.\(\frac{2x}{x^2-1}\)C.\(\frac{2}{x^2+1}\)D.\(\frac{2x}{x^2+1}\)6.下列分式中，分子和分母没有公因数的是（）A.\(\frac{3x}{6x}\)B.\(\frac{4x^2}{8x}\)C.\(\frac{5x}{10x+5}\)D.\(\frac{7x}{14x-7}\)7.如果\(x=2\)是分式\(\frac{2x-1}{x-1}\)的增根，那么这个分式的值为（）A.0B.1C.2D.无法确定8.计算\(\frac{1}{x-1}-\frac{1}{x+1}\)的结果是（）A.\(\frac{2}{x^2-1}\)B.\(\frac{2x}{x^2-1}\)C.\(\frac{2}{x^2+1}\)D.\(\frac{2x}{x^2+1}\)9.将分式\(\frac{3x+6}{2x-4}\)化简为最简分式，结果是（）A.\(\frac{3}{2}\)B.\(\frac{3x+6}{2x-4}\)C.\(\frac{3(x+2)}{2(x-2)}\)D.\(\frac{3}{2}\cdot\frac{x+2}{x-2}\)10.下列分式中，分母不能为零的是（）A.\(\frac{1}{x}\)B.\(\frac{1}{x+1}\)C.\(\frac{1}{x-1}\)D.以上都是二、填空题（每题3分，共30分）11.将分式\(\frac{2x^2-4x}{x^2-4}\)化简为最简分式，结果是\(\frac{2x}{x+2}\)。12.如果\(\frac{a}{b}=2\)，那么\(b=\frac{a}{2}\)。13.计算\(\frac{1}{x-2}\cdot\frac{x+2}{x+1}\)的结果是\(\frac{x+2}{x^2-x-2}\)。14.如果\(x=-1\)是分式\(\frac{x^2+1}{x+1}\)的增根，那么这个分式的值为\(0\)。15.将分式\(\frac{4x^2-4}{x^2-4x+4}\)化简为最简分式，结果是\(4\)。16.如果\(\frac{a}{b}=\frac{c}{d}\)，那么\(\frac{a+c}{b+d}=\frac{a}{b}\)。17.计算\(\frac{1}{x-1}+\frac{1}{x+1}+\frac{1}{x^2-1}\)的结果是\(\frac{x}{x^2-1}\)。18.将分式\(\frac{5x-5}{x^2-9}\)化简为最简分式，结果是\(\frac{5}{x+3}\)。19.如果\(x=2\)是分式\(\frac{2x-1}{x-1}\)的增根，那么这个分式的值为\(2\)。20.将分式\(\frac{6x^2-9x}{3x^2-6x}\)化简为最简分式，结果是\(2x-3\)。三、解答题（每题10分，共40分）21.计算\(\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x^2-1}\)，并化简结果。解：原式=\(\frac{(x+1)-(x-1)}{(x-1)(x+1)}+\frac{1}{x^2-1}\)=\(\frac{2}{x^2-1}+\frac{1}{x^2-1}\)=\(\frac{3}{x^2-1}\)=\(\frac{3}{(x+1)(x-1)}\)=\(\frac{3}{x^2-1}\)。22.已知\(\frac{a}{b}=\frac{c}{d}\)，求\(\frac{a+c}{b+d}\)的值。解：由已知\(\frac{a}{b}=\frac{c}{d}\)，得\(ad=bc\)。则\(\frac{a+c}{b+d}=\frac{a}{b}\cdot\frac{b+d}{b+d}+\frac{c}{d}\cdot\frac{b}{b+d}=\frac{ad+bc}{bd+bd}=\frac{ad+bc}{2bd}=\frac{2ad}{2bd}=\frac{a}{b}\)。23.计算\(\frac{2x}{x-1}-\frac{3}{x+1}+\frac{5}{x^2-1}\)，并化简结果。解：原式=\(\frac{2x(x+1)-3(x-1)+5}{(x-1)(x+1)}\)=\(\frac{2x^2+2x-3x+3+5}{x^2-1}\)=\(\frac{2x^2-x+8}{x^2-1}\)=\(\frac{(2x-1)(x+4)}{(x+1)(x-1)}\)。24.已知\(\frac{a}{b}=\frac{c}{d}\)，求\(\frac{a-d}{b-c}\)的值。解：由已知\(\frac{a}{b}=\frac{c}{d}\)，得\(ad=bc\)。则\(\frac{a-d}{b-c}=\frac{a}{b}\cdot\frac{b-c}{b-c}-\frac{d}{c}\cdot\frac{b-c}{b-c}=\frac{ab-bc-bd+cd}{bc-cd}=\frac{ab-bc-bd+cd}{bc-cd}=\frac{a}{b}-\frac{d}{c}\)。答案：1.C2.B3.A4.B5.B6.D7.B8.A9.C10.D11.\(\frac{2x}{x+2}\)12.\(