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题目
选择题
- 方程\(x^2-10x+25=0\)有（\(\qquad\)）个根.
A.0个\(\qquad\qquad\)B.1个\(\qquad\qquad\)C.2个\(\qquad\qquad\)D.3个
- 以下关于\(x\)的方程中，有（\(\qquad\)）个是一元二次方程（\(m\)为参数）.
\(\qquad\)①\(x^2+x+1=0\)
\(\qquad\)②\(x+\Large\frac{1}{x}\normalsize=3\)
\(\qquad\)③\(mx^2=1\)
\(\qquad\)④\(x^2+\sqrt{m+2}\space x-3m=0(m\geq-2)\)
\(\qquad\)⑤\(\Large\frac{x^4}{x^2}\normalsize+2x+1=0\)
\(\qquad\)⑥\(m^3(x^2+1)+m^2(x+3)-m(4x^2-1)=0(m\ne 0)\)
A.2个\(\qquad\qquad\)B.1个\(\qquad\qquad\)C.3个\(\qquad\qquad\)D.5个
- 已知关于\(x\)的方程\(x^2+4x+a\)有两个实根\(x_1\),\(x_2\)，且\(2x_1-x_2=7\).求实数\(a\)的值（\(\qquad\)）.
A.6\(\qquad\qquad\)B.0\(\qquad\qquad\)C.-5\(\qquad\qquad\)D.无法确定.
- 下列多项式中恒\(>0\)的有（\(\qquad\)）个.
\(\qquad\)①\(x^2+x+1\)
\(\qquad\)②\(x^2-5x+\large\frac{24}{4}\)
\(\qquad\)③\((k+1)x^2-(1-k^2)x+\large\frac{k^3-k^2-k+1}{4}\normalsize(k\geq -1)\)
\(\qquad\)④\(x+3\sqrt{x-3}-2\quad(\sqrt{x-3}\text{有意义})\)
A.4个\(\qquad\qquad\)B.3个\(\qquad\qquad\)C.2个\(\qquad\qquad\)D.1个
- 下列选项中最小值最小的是（\(\qquad\)）.
A.\(x^2+\large\frac{2\sqrt{10}}{5}\normalsize x+1\) \(\quad\) B.\(25x^2+100x+109\) \(\quad\) C.\(-x^2-2x+5\) \(\quad\) D.\(x^2+2x+\large\frac{114}{113}\)
- 若\(\alpha^2-3\alpha+1=0,\beta^2+6\beta+4=0\),则\(\large\frac{\alpha}{\beta}=\)（\(\qquad\)）.
A.\(-\frac{1}{2}\) \(\qquad\qquad\) B. \(2\text{或}-\frac{1}{2}\) \(\qquad\qquad\) C.\(-\frac{3}{4}\) \(\qquad\qquad\) D.ABC都不对.
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填空题
- 解方程\(3x^2-5|x|-12=0\)。__________.
- 已知关于\(x\)的二次多项式\(4mx^2-12mx+11m+\frac{1}{2}\)的最小值为\(2\)，则\(m=\) __________.
- 已知关于\(x\)的多项式\((m-1)x^2+2x+1\)没有最小值，则m的取值范围是__________.
- 若\(a^2-13a+1=0\)，\(b^2-13b+1=0\),则\(\large\frac{a}{b}\normalsize+\large\frac{b}{a}=\) __________.
- 设\(x_1\)，\(x_2\)为方程\(x^2-5x+3=0\)的两根。则\(\Large \frac{{x_1}^3+{x_2}^3}{{x_1}^2+{x_2}^2}\normalsize=\) __________.
- 若\(a,b\)是关于\(x\)的方程\(x^2+(p-2)x+1=0\)的两实根，则\((a^2+ap+1)(b^2+bp+1)=\) __________.
- 解不等式\(5x^2-11x+\frac{11}{2}\leq0\)。__________.
- 关于\(x^2-(m+1)x+3=0\)的两根满足\(|x_1|-3|x_2|=1\),则\(m\)的值为__________.
- 若关于\(x\)的方程\((a^2-1)\large(\frac{x+1}{x})^2\normalsize-(2a+3)\large\frac{x+1}{x}\normalsize+1=0\)有一个实数解，则\(a=\) __________.
- 设\(\alpha,\beta\)是方程\(2x^2+3x-1=0\)的两实根.请构造一个整系数方程，其两根分别为\(\alpha+\frac{1}{\beta},\beta+\frac{1}{\alpha}\).__________.
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解答题
- 解方程：\(x^2+(5+4\sqrt3)x+18+10\sqrt3=0\).
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- 已知关于\(x\)的一元二次方程\(x^2-mx-2m=0\)有两实根\(x_1,x_2\)满足\({x_1}^2+{x_2}^2=12\).求\(m\)的值.
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- 已知\(10x^2-125x+100=0\)的根为\(a,b\),求代数式\(8a^3+8a^2b+10ab^3\)的值.
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- 已知\(m,n\)是方程\(x^2+8x+4=0\)的两根，则\(m^4+10m^3+6m^3n-5m^2-5mn+18\)的值是.
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- 解关于\(x\)的方程：\(ax(x-4)=-4b\).
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关于\(x\)的方程\(x^2-(2k-3)x+k^2+1=0\)有两个不相等的实根\(x_1\),\(x_2\)
(1) 求\(k\)的范围.
(2) 求证：\(x_1<0,x_2<0\).
(3) 若\(x_1x_2-|x_1|-|x_2|=6\)，求\(k\).
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答案
选择题答案
- C
- A
- C
- C
- D
- B
填空题答案
- \(x_{1,2}=\pm3\)
- \(\large\frac{3}{4}\)
- \(m\leq 1\)
- \(2\text{或}167\)
- \(\large\frac{80}{19}\)
- \(4\)
- \(\frac{11-\sqrt{11}}{10}\leq x\leq\frac{11+\sqrt{11}}{10}\)
- \(m=\large\frac{-2+2\sqrt{37}}{3}\normalsize\text{或}\large\frac{-4-2\sqrt{37}}{3}\)
- \(a=\pm1\text{或}-\large\frac{13}{12}\)
- \(2x^2-x-1=0\)
解答题答案
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