《2025年全程检测单元测试卷九年级数学全一册湘教版》

20. (12分)综合与实践.
【问题情境】为了解方程$(x^{2})^{2}-5x^{2}+4 = 0$,我们把$x^{2}$看作一个整体,然后设$y = x^{2}$,则原方程可化为$y^{2}-5y + 4 = 0$,解得$y_{1}=1,y_{2}=4$,则$x^{2}=1$或$x^{2}=4$.可得原方程的解为$x_{1}=1,x_{2}=-1,x_{3}=2,x_{4}=-2$.我们通常把这种解决问题的方法叫作换元法.
【问题发现】已知实数$m,n$满足$m^{2}-2m - 3 = 0,n^{2}-2n - 3 = 0$,且$m\neq n$,显然$m,n$是方程$x^{2}-2x - 3 = 0$的两个不相等的实数根,由根与系数的关系可知$m + n = 2,mn=-3$.
根据上述材料,解决以下问题.
【实践探究】(1)(4分)方程$x^{4}-11x^{2}+28 = 0$的解为x₁ = 2，x₂ = - 2，x₃ = √7，x₄ = -√7.
(2)(4分)已知实数$a,b(a\neq b)$分别满足$3a^{2}-7a + 2 = 0,3b^{2}-7b + 2 = 0$,则$ab$的值是2/3.
【拓展应用】(3)(4分)已知正实数$a,b$分别满足$a^{4}-14a^{2}+36 = 0,b^{4}-14b^{2}+36 = 0$,求$a + b$的值.

答案： 20．
(1)$x_{1} = 2$，$x_{2} = - 2$，$x_{3} = \sqrt{7}$，$x_{4} = - \sqrt{7}$ 【解析】设$y = x^{2}$，则原方程可化为$y^{2} - 11y + 28 = 0$．解得$y_{1} = 4$，$y_{2} = 7$，则$x^{2} = 4$或$x^{2} = 7$．$\therefore$原方程的解为$x_{1} = 2$，$x_{2} = - 2$，$x_{3} = \sqrt{7}$，$x_{4} = - \sqrt{7}$．
(2)$\frac{2}{3}$ 【解析】$a$，$b$分别满足$3a^{2} - 7a + 2 = 0$，$3b^{2} - 7b + 2 = 0$，则$a$，$b$可看作方程$3x^{2} - 7x + 2 = 0$的两个根，所以$ab = \frac{2}{3}$．
(3)令$a^{2} = p$，$b^{2} = q$，则得$p^{2} - 14p + 36 = 0$，$q^{2} - 14q + 36 = 0$．$\therefore p$，$q$是方程$x^{2} - 14x + 36 = 0$的两个根，由根与系数的关系得$p + q = 14$，$pq = 36$．$\therefore a^{2} + b^{2} = 14$，$a^{2}b^{2} = 36$．又$\because a$，$b$是正实数，$\therefore ab = 6$．$\therefore a + b = \sqrt{(a + b)^{2}} = \sqrt{a^{2} + b^{2} + 2ab} = \sqrt{14 + 12} = \sqrt{26}$．