Question

已知方程x²+根号下2*x-1=0有2个根x1、x2 求①x1²+x2²

还有②（x1-x2）²③x1负一次方+x2的负一次方希望解答一下，必有重谢

Answer 1

解：由韦达定理得
x1+x2= -根号下2
x1*x2 =-1
①x1²+x2²=x1²+x2²+2x1*x2-2x1*x2
=(x1+x2)²-2x1*x2=2-(-1)=4
②（x1-x2）²=x1²+x2²-2x1*x2=4-2(-1)=6
③x1负一次方+x2的负一次方=(x1+x2)/x1*x2
=根号下2

Answer 2

方程x²+根号下2*x-1=0有2个根x1、x2
∴x1x2=-1
x1+x2=-√2

①x1²+x2² =(x1+x2)²-2x1x2=2-(-2)=4

②（x1-x2）²=(x1+x2)²-4x1x2=2-(-4)=-6
③x1负一次方+x2的负一次方=(x1+x2)/x1x2=√2

Answer 3

x²+根号下2x-1=0
x1+x2=-根号下2
x1x2=-1
x1²+x2²=(x1+x2)²-2x1x2=2+2=4
(x1-x2)²=x1²+x2²-2x1x2=4+2=6
1/x1+1/x2=(x1+x2)/x1x2=根号下2

Answer 4

①x1²+x2²=（x1+x2）²-2x1x2 =（—根号下2）²-2乘以（-1）=4
②（x1-x2）²=（x1+x2）²-4x1x2 =（—根号下2）²-4乘以（-1）=6
③x1负一次方+x2的负一次方=（x1x2）分之（x1+x2）=2分之根号下2