题目已知一元二次方程2x2−x−2=02x^{2}-x-2=02x2−x−2=0的两根是x1x_{1}x1,x2x_{2}x2,求下列代数式的值.(1)(x1−x2)2(1){({x}_{1}-{x}_{2})}^{2}(1)(x1−x2)2；(2)(x1+1)(x2+1)(2)(x_{1}+1)(x_{2}+1)(2)(x1+1)(x2+1).知识点：根的判别式章节：方程与不等式 / 一元二次方程
