先看这样一道题:已知x_1、x_2是方程x~2+x-1=0的两个根,求代数式(x_1~2+2x_1-1)(x_2~2+2x_2-1)的值.大多数同学采用以下方法进行的:原式=(x_1x_2)~2+2x_1~2x_2-x_1~2+2x_1x_2~2+4x_1x_2-2x_1-x_2~2-2x_2+1=(x_1x_2)~2+2x_1x_2(x_1+x_2)-(x_1+x_2)~2+6x_1x_2-2(x_1+x_2)+1.再以x_1+x_2=-1,x_1x_2=-1代入,计算出结果为-1.由于上述变形较繁,容易出现失误.实际上,本题可利用方程根的定义轻松解决:因为x_1、x_2是方程x~2+x-1=0的两 Consider the following question: Given x_1, x_2 is the root of the equation x ~ 2 + x-1 = 0, find the value of the algebraic equation (x_1 ~ 2 + 2x_1-1) (x_2 ~ 2 + 2x_2-1). Most students use the following methods: Original = (x_1x_2) ~ 2 + 2x_1 ~ 2x_2-x_1 ~ 2 + 2x_1x_2 ~ 2 + 4x_1x_2-2x_1-x_2 ~ 2-2x_2 + 1 = (x_1x_2) ~ 2 + 2x_1x_2 x_1 + x_2) - (x_1 + x_2) ~ 2 + 6x_1x_2-2 (x_1 + x_2) +1, then substitute x_1 + x_2 = -1 and x_1x_2 = -1, and calculate the result as -1. In fact, the problem can be easily solved by the definition of the root of the equation: Because x_1, x_2 are two of the equation x ~ 2 + x-1 = 0