2025年增根方程競賽題庫本文借鑒了近年相關(guān)經(jīng)典試題創(chuàng)作而成，力求幫助考生深入理解測試題型，掌握答題技巧，提升應(yīng)試能力。一、選擇題（每題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.無增根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.無增根10.方程\(\sqrt{x^2-4}=x-2\)的解集是A.\(\{2\}\)B.\(\{4\}\)C.\(\{2,4\}\)D.\(\emptyset\)二、填空題（每題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、解答題（每題10分，共50分）1.解方程\(\sqrt{x+1}+\sqrt{x-1}=2\)。2.解方程\(\frac{x^2}{x-1}=x\)，并判斷是否有增根。3.解方程\(\sqrt{2x-1}+\sqrt{x+3}=4\)，并判斷是否有增根。4.解方程\(\sqrt{x^2-4}=x-2\)，并判斷是否有增根。5.解方程\(\frac{1}{x-1}+\frac{1}{x+1}=\frac{2}{x}\)，并判斷是否有增根。答案和解析一、選擇題1.C-解析：方程\(\frac{x^2-1}{x-1}=1\)化簡為\(x+1=1\)，解得\(x=0\)。同時，需排除\(x=1\)使分母為零的情況。因此，解集為\(\{0\}\)。2.A-解析：方程\(\sqrt{x+1}+\sqrt{x-1}=2\)兩邊平方得\(x+1+x-1+2\sqrt{(x+1)(x-1)}=4\)，化簡得\(2x+2\sqrt{x^2-1}=4\)，進一步化簡得\(x+\sqrt{x^2-1}=2\)，解得\(x=1\)。3.C-解析：方程\(\frac{1}{x-1}+\frac{1}{x+1}=\frac{2}{x}\)化簡為\(\frac{x+1+x-1}{(x-1)(x+1)}=\frac{2}{x}\)，即\(\frac{2x}{x^2-1}=\frac{2}{x}\)，解得\(x=0\)為增根。4.A-解析：方程\(\sqrt{x-1}=x-3\)兩邊平方得\(x-1=(x-3)^2\)，化簡得\(x-1=x^2-6x+9\)，即\(x^2-7x+10=0\)，解得\(x=2\)或\(x=5\)。檢驗發(fā)現(xiàn)\(x=5\)不滿足原方程，故解為\(x=2\)。5.B-解析：方程\(\frac{x^2}{x-1}=x\)化簡為\(x^2=x(x-1)\)，即\(x^2=x^2-x\)，解得\(x=0\)為增根，實際解為\(x=2\)。6.A-解析：方程\(\sqrt{x+2}-\sqrt{x-2}=0\)化簡為\(\sqrt{x+2}=\sqrt{x-2}\)，兩邊平方得\(x+2=x-2\)，解得\(x=2\)。7.B-解析：方程\(\frac{x^2-4}{x-2}=3\)化簡為\(x+2=3\)，解得\(x=1\)為增根，實際解為\(x=3\)。8.A-解析：方程\(\sqrt{2x-1}+\sqrt{x+3}=4\)兩邊平方得\(2x-1+x+3+2\sqrt{(2x-1)(x+3)}=16\)，化簡得\(3x+2+2\sqrt{2x^2+5x-3}=16\)，進一步化簡得\(2\sqrt{2x^2+5x-3}=14-3x\)，解得\(x=3\)。9.A-解析：方程\(\frac{x^2+1}{x-1}=x+1\)化簡為\(x^2+1=(x+1)(x-1)\)，即\(x^2+1=x^2-1\)，解得\(x=1\)為增根。10.B-解析：方程\(\sqrt{x^2-4}=x-2\)兩邊平方得\(x^2-4=(x-2)^2\)，化簡得\(x^2-4=x^2-4x+4\)，解得\(x=4\)。二、填空題1.1-解析：方程\(\sqrt{x-1}+\sqrt{x+1}=\sqrt{2x+2}\)兩邊平方得\(x-1+x+1+2\sqrt{(x-1)(x+1)}=2x+2\)，化簡得\(2x+2\sqrt{x^2-1}=2x+2\)，進一步化簡得\(\sqrt{x^2-1}=1\)，解得\(x=1\)。2.2-解析：方程\(\frac{x^2-1}{x-1}=2\)化簡為\(x+1=2\)，解得\(x=1\)為增根，實際解為\(x=2\)。3.3-解析：方程\(\sqrt{x+3}-\sqrt{x+1}=1\)兩邊平方得\(x+3+x+1-2\sqrt{(x+3)(x+1)}=1\)，化簡得\(2x+4-2\sqrt{x^2+4x+3}=1\)，進一步化簡得\(2\sqrt{x^2+4x+3}=2x+3\)，解得\(x=3\)。4.1-解析：方程\(\frac{1}{x-1}+\frac{1}{x+1}=\frac{2}{x}\)化簡為\(\frac{x+1+x-1}{(x-1)(x+1)}=\frac{2}{x}\)，即\(\frac{2x}{x^2-1}=\frac{2}{x}\)，解得\(x=0\)為增根。5.2-解析：方程\(\sqrt{x-2}+\sqrt{x-3}=1\)兩邊平方得\(x-2+x-3+2\sqrt{(x-2)(x-3)}=1\)，化簡得\(2x-5+2\sqrt{x^2-5x+6}=1\)，進一步化簡得\(2\sqrt{x^2-5x+6}=6-2x\)，解得\(x=2\)。三、解答題1.解方程\(\sqrt{x+1}+\sqrt{x-1}=2\)-解：兩邊平方得\(x+1+x-1+2\sqrt{(x+1)(x-1)}=4\)，化簡得\(2x+2\sqrt{x^2-1}=4\)，進一步化簡得\(x+\sqrt{x^2-1}=2\)，解得\(x=1\)。2.解方程\(\frac{x^2}{x-1}=x\)，并判斷是否有增根-解：方程化簡為\(x^2=x(x-1)\)，即\(x^2=x^2-x\)，解得\(x=0\)為增根，實際解為\(x=2\)。3.解方程\(\sqrt{2x-1}+\sqrt{x+3}=4\)，并判斷是否有增根-解：兩邊平方得\(2x-1+x+3+2\sqrt{(2x-1)(x+3)}=16\)，化簡得\(3x+2+2\sqrt{2x^2+5x-3}=16\)，進一步化簡得\(2\sqrt{2x^2+5x-3}=14-3x\)，解得\(x=3\)。4.解方程\(\sqrt{x^2-4}=x-2\)，并判斷是否有增根-解：兩邊平方得\(x^2-4=(x-2)^2\)，