设X1、X2是一元二次方程2X^2-5X+1=0的两个根，利用根与系数的关系,求下列各式的值 。

解：
x1x2=0.5 x1+x2=2.5

1、(x1-3)(x2-3)
=x1x2-3(x1+x2)+9
=0.5-7.5+9
=2
2、(x1+1)²+(x2+1)²
=x1²+2x1+1+x2²+2x2+1
=(x1²+x2²)+2(x1+x2)+2
=(x1+x2)²-2x1x2+2(x1+x2)+2
=6.25-1+5+2
=12.25
3、x2/x1+x1/x2
=(x2²+x1²)/x1x2
=[(x1+x2)²-2x1x2]/x1x2
=(6.25-1)/0.5
=10.5
4、|x1-x2|²=(x1-x2)²
=(x1+x2)²-4x1x2
=6.25-2
=4.25
所以：|x1-x2|=√4.25=√17/2

x1+x2=-b/a=5/2 x1*x2=c/a=1/2 (x1-x2)^2=x1^2-2x1*x2+x2^2=x1^2+2x1*x2+x2^2-4x1*x2=(x1+x2)^2-4x1*x2=(5/2)^2-4*1/2=25/4-2=17/4 |x1-x2|=根号下17/4==根号17/2