Question

设x1,x2是一元二次方程2x2-5x+1=0的两个根,利用根与系数的关系,求下列各式的值

（x1+1）²+(x2+1)²=
x1³+x2³=
x2／x1+x1 + x1／x2+1=
(x1+ 1／3x2)(x2+ 1／3x1)=

Answer 1

（x1+1/3x2)(x2+1/3x1)
=x1*x2+(x1+x2)的平方/3-2x1*x2/3+x1*x2/3
=(x1+x2)的平方/3+2x1*x2/3
=4/27-4/9
=-8/27

Answer 2

（x1+1）²+(x2+1)²=(x1+x2)2-2x1x2+2(x1+x2)+2
x1³+x2³=(x1+x2)((x1+x2)2-3x1x2)
x2／x1+x1 + x1／x2+1、部分式子引入根式
(x1+ 1／3x2)(x2+ 1／3x1)=芳 上

Answer 3

由韦达定理得：x1 + x2 = 2/5
x1 x2 =2
（x1+1）²+(x2+1)²= x1² +x2² +2（x1+x2）+ 2 = (x1 + x2 )² - 2 x1 x2 + 2（x1+x2）+ 2= -(26/25)