题目已知aaa为正实数,x1x_{1}x1,x2x_{2}x2是方程x2−ax−a=0x^{2}-ax-a=0x2−ax−a=0的两个根,则（x12−1)(x22−1)=( ){x}_{1}^{2}-1)({x}_{2}^{2}-1)=\left(\ \ \right)x12−1)(x22−1)=( )A.2a+12a+12a+1B.2a−12a-12a−1C.−2a+1-2a+1−2a+1D.−2a−1-2a-1−2a−1知识点：解一元二次方程——因式分解法章节：方程与不等式 / 一元二次方程
