2025年增根方程竞赛题库

2025年增根方程竞赛题库本文借鉴了近年相关经典试题创作而成，力求帮助考生深入理解测试题型，掌握答题技巧，提升应试能力。一、选择题（每题3分，共30分）1.方程\(\frac{x^2-1}{x-1}=1\)的解集是A.\(\{0\}\)B.\(\{2\}\)C.\(\{0,2\}\)D.\(\emptyset\)2.方程\(\sqrt{x+1}+\sqrt{x-1}=2\)的解集是A.\(\{1\}\)B.\(\{2\}\)C.\(\{1,2\}\)D.\(\emptyset\)3.方程\(\frac{1}{x-1}+\frac{1}{x+1}=\frac{2}{x}\)的增根是A.\(x=1\)B.\(x=-1\)C.\(x=0\)D.无增根4.方程\(\sqrt{x-1}=x-3\)的解集是A.\(\{2\}\)B.\(\{3\}\)C.\(\{2,3\}\)D.\(\emptyset\)5.方程\(\frac{x^2}{x-1}=x\)的解集是A.\(\{1\}\)B.\(\{2\}\)C.\(\{1,2\}\)D.\(\emptyset\)6.方程\(\sqrt{x+2}-\sqrt{x-2}=0\)的解集是A.\(\{2\}\)B.\(\{4\}\)C.\(\{2,4\}\)D.\(\emptyset\)7.方程\(\frac{x^2-4}{x-2}=3\)的解集是A.\(\{1\}\)B.\(\{3\}\)C.\(\{1,3\}\)D.\(\emptyset\)8.方程\(\sqrt{2x-1}+\sqrt{x+3}=4\)的解集是A.\(\{3\}\)B.\(\{4\}\)C.\(\{3,4\}\)D.\(\emptyset\)9.方程\(\frac{x^2+1}{x-1}=x+1\)的增根是A.\(x=1\)B.\(x=-1\)C.\(x=0\)D.无增根10.方程\(\sqrt{x^2-4}=x-2\)的解集是A.\(\{2\}\)B.\(\{4\}\)C.\(\{2,4\}\)D.\(\emptyset\)二、填空题（每题4分，共20分）1.方程\(\sqrt{x-1}+\sqrt{x+1}=\sqrt{2x+2}\)的解是\(x=\)。2.方程\(\frac{x^2-1}{x-1}=2\)的解是\(x=\)。3.方程\(\sqrt{x+3}-\sqrt{x+1}=1\)的解是\(x=\)。4.方程\(\frac{1}{x-1}+\frac{1}{x+1}=\frac{2}{x}\)的增根是\(x=\)。5.方程\(\sqrt{x-2}+\sqrt{x-3}=1\)的解是\(x=\)。三、解答题（每题10分，共50分）1.解方程\(\sqrt{x+1}+\sqrt{x-1}=2\)。2.解方程\(\frac{x^2}{x-1}=x\)，并判断是否有增根。3.解方程\(\sqrt{2x-1}+\sqrt{x+3}=4\)，并判断是否有增根。4.解方程\(\sqrt{x^2-4}=x-2\)，并判断是否有增根。5.解方程\(\frac{1}{x-1}+\frac{1}{x+1}=\frac{2}{x}\)，并判断是否有增根。答案和解析一、选择题1.C-解析：方程\(\frac{x^2-1}{x-1}=1\)化简为\(x+1=1\)，解得\(x=0\)。同时，需排除\(x=1\)使分母为零的情况。因此，解集为\(\{0\}\)。2.A-解析：方程\(\sqrt{x+1}+\sqrt{x-1}=2\)两边平方得\(x+1+x-1+2\sqrt{(x+1)(x-1)}=4\)，化简得\(2x+2\sqrt{x^2-1}=4\)，进一步化简得\(x+\sqrt{x^2-1}=2\)，解得\(x=1\)。3.C-解析：方程\(\frac{1}{x-1}+\frac{1}{x+1}=\frac{2}{x}\)化简为\(\frac{x+1+x-1}{(x-1)(x+1)}=\frac{2}{x}\)，即\(\frac{2x}{x^2-1}=\frac{2}{x}\)，解得\(x=0\)为增根。4.A-解析：方程\(\sqrt{x-1}=x-3\)两边平方得\(x-1=(x-3)^2\)，化简得\(x-1=x^2-6x+9\)，即\(x^2-7x+10=0\)，解得\(x=2\)或\(x=5\)。检验发现\(x=5\)不满足原方程，故解为\(x=2\)。5.B-解析：方程\(\frac{x^2}{x-1}=x\)化简为\(x^2=x(x-1)\)，即\(x^2=x^2-x\)，解得\(x=0\)为增根，实际解为\(x=2\)。6.A-解析：方程\(\sqrt{x+2}-\sqrt{x-2}=0\)化简为\(\sqrt{x+2}=\sqrt{x-2}\)，两边平方得\(x+2=x-2\)，解得\(x=2\)。7.B-解析：方程\(\frac{x^2-4}{x-2}=3\)化简为\(x+2=3\)，解得\(x=1\)为增根，实际解为\(x=3\)。8.A-解析：方程\(\sqrt{2x-1}+\sqrt{x+3}=4\)两边平方得\(2x-1+x+3+2\sqrt{(2x-1)(x+3)}=16\)，化简得\(3x+2+2\sqrt{2x^2+5x-3}=16\)，进一步化简得\(2\sqrt{2x^2+5x-3}=14-3x\)，解得\(x=3\)。9.A-解析：方程\(\frac{x^2+1}{x-1}=x+1\)化简为\(x^2+1=(x+1)(x-1)\)，即\(x^2+1=x^2-1\)，解得\(x=1\)为增根。10.B-解析：方程\(\sqrt{x^2-4}=x-2\)两边平方得\(x^2-4=(x-2)^2\)，化简得\(x^2-4=x^2-4x+4\)，解得\(x=4\)。二、填空题1.1-解析：方程\(\sqrt{x-1}+\sqrt{x+1}=\sqrt{2x+2}\)两边平方得\(x-1+x+1+2\sqrt{(x-1)(x+1)}=2x+2\)，化简得\(2x+2\sqrt{x^2-1}=2x+2\)，进一步化简得\(\sqrt{x^2-1}=1\)，解得\(x=1\)。2.2-解析：方程\(\frac{x^2-1}{x-1}=2\)化简为\(x+1=2\)，解得\(x=1\)为增根，实际解为\(x=2\)。3.3-解析：方程\(\sqrt{x+3}-\sqrt{x+1}=1\)两边平方得\(x+3+x+1-2\sqrt{(x+3)(x+1)}=1\)，化简得\(2x+4-2\sqrt{x^2+4x+3}=1\)，进一步化简得\(2\sqrt{x^2+4x+3}=2x+3\)，解得\(x=3\)。4.1-解析：方程\(\frac{1}{x-1}+\frac{1}{x+1}=\frac{2}{x}\)化简为\(\frac{x+1+x-1}{(x-1)(x+1)}=\frac{2}{x}\)，即\(\frac{2x}{x^2-1}=\frac{2}{x}\)，解得\(x=0\)为增根。5.2-解析：方程\(\sqrt{x-2}+\sqrt{x-3}=1\)两边平方得\(x-2+x-3+2\sqrt{(x-2)(x-3)}=1\)，化简得\(2x-5+2\sqrt{x^2-5x+6}=1\)，进一步化简得\(2\sqrt{x^2-5x+6}=6-2x\)，解得\(x=2\)。三、解答题1.解方程\(\sqrt{x+1}+\sqrt{x-1}=2\)-解：两边平方得\(x+1+x-1+2\sqrt{(x+1)(x-1)}=4\)，化简得\(2x+2\sqrt{x^2-1}=4\)，进一步化简得\(x+\sqrt{x^2-1}=2\)，解得\(x=1\)。2.解方程\(\frac{x^2}{x-1}=x\)，并判断是否有增根-解：方程化简为\(x^2=x(x-1)\)，即\(x^2=x^2-x\)，解得\(x=0\)为增根，实际解为\(x=2\)。3.解方程\(\sqrt{2x-1}+\sqrt{x+3}=4\)，并判断是否有增根-解：两边平方得\(2x-1+x+3+2\sqrt{(2x-1)(x+3)}=16\)，化简得\(3x+2+2\sqrt{2x^2+5x-3}=16\)，进一步化简得\(2\sqrt{2x^2+5x-3}=14-3x\)，解得\(x=3\)。4.解方程\(\sqrt{x^2-4}=x-2\)，并判断是否有增根-解：两边平方得\(x^2-4=(x-2)^2\)，