20. 核心素养运算能力「2026四川绵阳北川期中，★☆」(8分)已知关于x的一元二次方程$x^2-(m+2)x+m-1=0$。
(1)求证:无论m取何值,方程都有两个不相等的实数根.
(2)如果方程的两个实数根为$x_1,x_2$,且$x_1^2+x_2^2 -x_1x_2=9$,求m的值.

解析 (1)证明:$\because a=1$,$b=-(m+2)$,$c=m-1$,
$\therefore \Delta=b^2-4ac=[-(m+2)]^2-4×1×(m-1)$
$=m^2+4m+4-4m+4=m^2+8$. $·············································$ (2分)
$\because m^2≥0$,$\therefore \Delta＞0$.$\therefore$ 无论$m$取何值,方程都有两个不相等的实数根.$·$ (4分)
(2)$\because$ 方程$x^2-(m+2)x+m-1=0$的两个实数根为$x_1,x_2$,
$\therefore x_1+x_2=m+2$,$x_1x_2=m-1$. $·$ (5分)
$\because x_1^2+x_2^2-x_1x_2=9$,$\therefore (x_1+x_2)^2-3x_1x_2=9$, $·································$ (7分)
$\therefore (m+2)^2-3(m-1)=9$.整理,得$m^2+m-2=0$.
$\therefore (m+2)(m-1)=0$.解得$m_1=-2$,$m_2=1$.
$\therefore m$的值为$-2$或$1$.$·$ (8分)