2025年事業(yè)單位招聘考試教師招聘數學學科專業(yè)知識試卷（代數）考試時間：______分鐘總分：______分姓名：______一、單項選擇題（本大題共10小題，每小題2分，共20分。每小題給出的四個選項中，只有一項是符合題目要求的。）1.設函數\(f(x)=x^3-3x^2+4x-1\)，則\(f(x)\)的極值點為：A.\(x=-1\)B.\(x=1\)C.\(x=2\)D.\(x=3\)2.若\(\log_2(x-1)+\log_2(x-2)=1\)，則\(x\)的值為：A.3B.4C.5D.63.設\(a,b\)是方程\(x^2-2ax+1=0\)的兩個實數根，則\(a+b\)的值為：A.0B.2C.4D.64.若\(\frac{1}{a}+\frac{1}=\frac{4}{ab}\)，則\(a^2+b^2\)的值為：A.8B.10C.12D.145.已知\(a,b,c\)成等差數列，且\(a^2+b^2+c^2=3\)，則\(ab+bc+ca\)的值為：A.0B.1C.2D.36.若\(\sin^2x+\cos^2x=1\)，則\(\tan^2x+\cot^2x\)的值為：A.2B.3C.4D.57.已知\(\log_2(x-1)-\log_2(x-2)=1\)，則\(x\)的值為：A.2B.3C.4D.58.設\(a,b\)是方程\(x^2-4x+3=0\)的兩個實數根，則\(ab\)的值為：A.1B.2C.3D.49.若\(\frac{1}{a}-\frac{1}=\frac{2}{ab}\)，則\(a^2+b^2\)的值為：A.2B.4C.6D.810.已知\(a,b,c\)成等比數列，且\(a^2+b^2+c^2=12\)，則\(ab+bc+ca\)的值為：A.2B.4C.6D.8二、填空題（本大題共5小題，每小題3分，共15分。把答案填在題中的橫線上。）1.若\(f(x)=2x^3-3x^2+4x-1\)，則\(f'(x)=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\三、解答題（本大題共2小題，共20分。請將答案寫在答題卡相應的位置上。）3.設\(f(x)=x^3-6x^2+9x-1\)，求\(f(x)\)的導數\(f'(x)\)，并找出\(f(x)\)的極值點。要求：首先，使用導數的定義求出\(f'(x)\)。然后，解方程\(f'(x)=0\)找出\(f(x)\)的極值點。最后，通過計算\(f(x)\)在極值點的值，確定極值點處的極大值或極小值。四、應用題（本大題共1小題，共20分。請將答案寫在答題卡相應的位置上。）4.某學校舉辦數學競賽，共有100名學生參加。已知參加競賽的學生中，有60%的學生參加了數學題，有70%的學生參加了物理題，有50%的學生參加了化學題。又有30%的學生同時參加了數學和物理題，有20%的學生同時參加了物理和化學題，有10%的學生同時參加了數學和化學題，且有5%的學生同時參加了數學、物理和化學題。要求：計算至少參加了數學、物理或化學題中的兩種題目的學生人數。本次試卷答案如下：一、單項選擇題（本大題共10小題，每小題2分，共20分。）1.B.\(x=1\)解析：首先，求函數\(f(x)=x^3-3x^2+4x-1\)的導數\(f'(x)=3x^2-6x+4\)。然后，令\(f'(x)=0\)解得\(x=1\)或\(x=\frac{2}{3}\)。通過檢驗，發(fā)現\(x=1\)是極大值點。2.B.4解析：根據對數函數的性質，將等式\(\log_2(x-1)+\log_2(x-2)=1\)轉化為\(\log_2[(x-1)(x-2)]=1\)，進一步得到\((x-1)(x-2)=2\)。解得\(x=3\)或\(x=4\)，通過檢驗，發(fā)現\(x=4\)滿足原方程。3.B.2解析：根據一元二次方程的根與系數的關系，有\(zhòng)(a+b=-\frac{a}\)。對于方程\(x^2-2ax+1=0\)，系數\(a=-2\)，\(b=1\)，所以\(a+b=-\frac{1}{-2}=2\)。4.A.8解析：根據題意，\(\frac{1}{a}+\frac{1}=\frac{4}{ab}\)可以轉化為\(\frac{a+b}{ab}=\frac{4}{ab}\)，進一步得到\(a+b=4\)。由\((a+b)^2=a^2+b^2+2ab\)，代入\(a+b=4\)得\(16=a^2+b^2+2ab\)，化簡得\(a^2+b^2=8\)。5.A.0解析：因為\(a,b,c\)成等差數列，所以\(b=a+d\)，\(c=a+2d\)。代入\(a^2+b^2+c^2=3\)得\(3a^2+6ad+5d^2=3\)，化簡得\(a^2+2ad+\frac{5}{3}d^2=1\)。又因為\(a+b+c=3a+3d=3\)，所以\(d=1-\frac{a}{3}\)。代入\(a^2+2ad+\frac{5}{3}d^2=1\)得\(a^2+2a(1-\frac{a}{3})+\frac{5}{3}(1-\frac{a}{3})^2=1\)，化簡得\(4a^2-6a+2=0\)，解得\(a=\frac{1}{2}\)或\(a=\frac{1}{2}\)。因此\(b=a+d=\frac{1}{2}+\frac{1}{2}=1\)，\(c=a+2d=\frac{1}{2}+\frac{2}{2}=\frac{3}{2}\)。所以\(ab+bc+ca=1\times1+1\times\frac{3}{2}+\frac{3}{2}\times\frac{1}{2}=0\)。6.B.3解析：由三角恒等式\(\sin^2x+\cos^2x=1\)，得到\(\tan^2x+\cot^2x=\frac{\sin^2x}{\cos^2x}+\frac{\cos^2x}{\sin^2x}=\frac{\sin^4x+\cos^4x}{\sin^2x\cos^2x}\)。利用恒等式\(\sin^2x+\cos^2x=1\)和\(\sin^2x\cos^2x=\frac{1}{4}\sin^2(2x)\)，得\(\tan^2x+\cot^2x=\frac{(1-\cos^2x)^2+\cos^4x}{\frac{1}{4}\sin^2(2x)}=\frac{1+2\cos^2x+\cos^4x}{\frac{1}{4}\sin^2(2x)}\)。由于\(\sin^2(2x)=4\sin^2x\cos^2x\)，得\(\tan^2x+\cot^2x=\frac{1+2\cos^2x+\cos^4x}{\sin^2(2x)}=\frac{1+2\cos^2x+\cos^4x}{4\sin^2x\cos^2x}=\frac{1+2\cos^2x+\cos^4x}{4(1-\cos^2x)(1+\cos^2x)}\)。化簡得\(\tan^2x+\cot^2x=3\)。7.B.3解析：同樣地，將等式\(\log_2(x-1)-\log_2(x-2)=1\)轉化為\(\log_2\frac{x-1}{x-2}=1\)，進一步得到\(\frac{x-1}{x-2}=2\)。解得\(x=3\)或\(x=4\)，通過檢驗，發(fā)現\(x=3\)滿足原方程。8.A.1解析：根據一元二次方程的根與系數的關系，有\(zhòng)(ab=\frac{c}{a}\)。對于方程\(x^2-4x+3=0\)，系數\(a=1\)，\(b=-4\)，\(c=3\)，所以\(ab=\frac{3}{1}=3\)。9.C.6解析：根據題意，\(\frac{1}{a}-\frac{1}=\frac{2}{ab}\)可以轉化為\(\frac{b-a}{ab}=\frac{2}{ab}\)，進一步得到\(b-a=2\)。由\((a-b)^2=a^2+b^2-2ab\)，代入\(b-a=2\)得\(4=a^2+b^2-2ab\)，化簡得\(a^2+b^2=6\)。10.C.6解析：因為\(a,b,c\)成等比數列，所以\(b^2=ac\)。代入\(a^2+b^2+c^2=12\)得\(a^2+ac+c^2=12\)。由于\((a+c)^2=a^2+2ac+c^2\)，得\((a+c)^2=12+ac\)。又因為\(a+b+c=3a+3d=3\)，所以\(a+c=3-b\)。代入\((a+c)^2=12+ac\)得\((3-b)^2=12+b^2\)，化簡得\(9-6b+b^2=12+b^2\)，解得\(b=1\)。因此\(a+c=3-1=2\)，\(ac=b^2=1\)，所以\(ab+bc+ca=1+1+1=3\)。二、填空題（本大題共5小題，每小題3分，共15分。）1.\(f'(x)=6x^2-6x+4\)解析：使用導數的定義，對\(f(x)=2x^3-3x^2+4x-1\)進行求導，得到\(f'(x)=6x^2-6x+4\)。2.\(x=4\)解析：根據對數函數的性質，將等式\(\log_2(x-1)+\log_2(x-2)=1\)轉化為\(\log_2[(x-1)(x-2)]=1\)，進一步得到\((x-1)(x-2)=2\)。解得\(x=3\)或\(x=4\)，通過檢驗，發(fā)現\(x=4\)滿足原方程。3.\(a+b=2\)解析：根據一元二次方程的根與系數的關系，有\(zhòng)(a+b=-\frac{a}\)。對于方程\(x^2-2ax+1=0\)，系數\(a=-2\)，\(b=1\)，所以\(a+b=-\frac{1}{-2}=2\)。4.\(a^2+b^2=8\)解析：根據題意，\(\frac{1}{a}+\frac{1}=\frac{4}{ab}\)可以轉化為\(\frac{a+b}{ab}=\frac{4}{ab}\)，進一步得到\(a+b=4\)。由\((a+b)^2=a^2+