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10.(2025·广安)已知方程$x^2 -5x -24=0$的两根分别为a和b，则代数式$a^2 -4a +b$的值为
29
.答案
10.29
11.已知实数$m,n$满足$m^2 + 2m -1=0$，$n^2 + 2n -1=0$，且$m≠n$，则$\frac{1}{m}+\frac{1}{n}=$
2
.答案
11.2
12.(2025·宿迁泗洪县期中)已知关于$x$的一元二次方程$x^2 - px + 1 = 0$($p$为常数)有两个不相等的实数根$x_1$和$x_2$.
(1)求下列代数式的值:①$\frac{1}{x_1} + \frac{1}{x_2}$；②$\frac{x_1}{x_2} + \frac{x_2}{x_1}$.
(2)已知$x_1^2 + x_2^2 = 2p + 1$，求$p$的值.
(1)求下列代数式的值:①$\frac{1}{x_1} + \frac{1}{x_2}$；②$\frac{x_1}{x_2} + \frac{x_2}{x_1}$.
(2)已知$x_1^2 + x_2^2 = 2p + 1$，求$p$的值.
答案
12.解:(1)由根与系数的关系,得 $x_1 + x_2 = p$,$x_1x_2=1$,
∴①原式$=\frac{x_1 + x_2}{x_1x_2}=\frac{p}{1}=p$.
②原式$=\frac{x_1^2 + x_2^2}{x_1x_2}=\frac{(x_1 + x_2)^2 - 2x_1x_2}{x_1x_2}=\frac{p^2 - 2×1}{1}=p^2 - 2$.
(2)$\because x_1^2 + x_2^2 = 2p + 1$,$\therefore (x_1 + x_2)^2 - 2x_1x_2 = 2p + 1$,
$\therefore p^2 - 2 = 2p + 1$,$\therefore p^2 - 2p - 3 = 0$,
$\therefore p_1=3$,$p_2=-1$.
$\because$ 当 $p=-1$ 时,方程 $x^2 + x + 1 = 0$ 无实根,
$\therefore p_2=-1$ 舍去,从而 $p=3$.
综上,$p$ 的值为 3.
∴①原式$=\frac{x_1 + x_2}{x_1x_2}=\frac{p}{1}=p$.
②原式$=\frac{x_1^2 + x_2^2}{x_1x_2}=\frac{(x_1 + x_2)^2 - 2x_1x_2}{x_1x_2}=\frac{p^2 - 2×1}{1}=p^2 - 2$.
(2)$\because x_1^2 + x_2^2 = 2p + 1$,$\therefore (x_1 + x_2)^2 - 2x_1x_2 = 2p + 1$,
$\therefore p^2 - 2 = 2p + 1$,$\therefore p^2 - 2p - 3 = 0$,
$\therefore p_1=3$,$p_2=-1$.
$\because$ 当 $p=-1$ 时,方程 $x^2 + x + 1 = 0$ 无实根,
$\therefore p_2=-1$ 舍去,从而 $p=3$.
综上,$p$ 的值为 3.
13.已知关于$x$的一元二次方程$x^2 - 2x + a = 0$的两个实数根$x_1, x_2$满足$x_1x_2 + x_1 + x_2 ＞ 0$，求$a$的取值范围。
答案
13.解:$\because x_1$,$x_2$ 是 $x^2 - 2x + a = 0$ 的两个实数根,
$\therefore x_1 + x_2 = 2$,$x_1x_2 = a$.
$\because x_1x_2 + x_1 + x_2 ＞ 0$,$\therefore 2 + a ＞ 0$,解得 $a ＞ -2$.
又$\because$ 根的判别式$(-2)^2 - 4a ≥ 0$,解得 $a ≤ 1$,
$\therefore a$ 的取值范围为 $-2 ＜ a ≤ 1$.
$\therefore x_1 + x_2 = 2$,$x_1x_2 = a$.
$\because x_1x_2 + x_1 + x_2 ＞ 0$,$\therefore 2 + a ＞ 0$,解得 $a ＞ -2$.
又$\because$ 根的判别式$(-2)^2 - 4a ≥ 0$,解得 $a ≤ 1$,
$\therefore a$ 的取值范围为 $-2 ＜ a ≤ 1$.
14.(2025·南充)设$x_1,x_2$是关于$x$的方程$(x-1)(x-2)=m^2$的两根.
(1)当$x_1=-1$时,求$x_2$及$m$的值.
(2)求证:$(x_1-1)(x_2-1)≤0$.
(1)当$x_1=-1$时,求$x_2$及$m$的值.
(2)求证:$(x_1-1)(x_2-1)≤0$.
答案
14.(1)解:把 $x_1=-1$ 代入方程$(x-1)(x-2)=m^2$,
得 $m^2=6$,
$\therefore m=\pm\sqrt{6}$,$\therefore (x-1)(x-2)=6$,即 $x^2 - 3x - 4 = 0$,
$\therefore (x-4)(x+1)=0$,$\therefore x_1=-1$,$x_2=4$,
$\therefore x_2=4$,$m=\pm\sqrt{6}$.
(2)证明:$\because$ 方程$(x-1)(x-2)=m^2$ 即 $x^2 - 3x + 2 - m^2=0$ 的两根为 $x_1$,$x_2$,
$\therefore x_1 + x_2 = 3$,$x_1 · x_2 = 2 - m^2$,
$\therefore (x_1 - 1)(x_2 - 1)=x_1 · x_2 - (x_1 + x_2) + 1=2 - m^2 - 3 + 1=-m^2$.
$\because m^2 ≥ 0$,$\therefore -m^2 ≤ 0$,即$(x_1 - 1)(x_2 - 1) ≤ 0$.
得 $m^2=6$,
$\therefore m=\pm\sqrt{6}$,$\therefore (x-1)(x-2)=6$,即 $x^2 - 3x - 4 = 0$,
$\therefore (x-4)(x+1)=0$,$\therefore x_1=-1$,$x_2=4$,
$\therefore x_2=4$,$m=\pm\sqrt{6}$.
(2)证明:$\because$ 方程$(x-1)(x-2)=m^2$ 即 $x^2 - 3x + 2 - m^2=0$ 的两根为 $x_1$,$x_2$,
$\therefore x_1 + x_2 = 3$,$x_1 · x_2 = 2 - m^2$,
$\therefore (x_1 - 1)(x_2 - 1)=x_1 · x_2 - (x_1 + x_2) + 1=2 - m^2 - 3 + 1=-m^2$.
$\because m^2 ≥ 0$,$\therefore -m^2 ≤ 0$,即$(x_1 - 1)(x_2 - 1) ≤ 0$.
