设x1和x2是方程2x²-6x+3=0,求下列各式的值.x1²x2+x1x2² 1/x1-1/x2 x1-x2用韦达定理怎么变形?

来源:学生作业帮助网 编辑:作业帮 时间:2024/03/29 14:05:41
设x1和x2是方程2x²-6x+3=0,求下列各式的值.x1²x2+x1x2² 1/x1-1/x2 x1-x2用韦达定理怎么变形?

设x1和x2是方程2x²-6x+3=0,求下列各式的值.x1²x2+x1x2² 1/x1-1/x2 x1-x2用韦达定理怎么变形?
设x1和x2是方程2x²-6x+3=0,求下列各式的值.
x1²x2+x1x2² 1/x1-1/x2
x1-x2用韦达定理怎么变形?

设x1和x2是方程2x²-6x+3=0,求下列各式的值.x1²x2+x1x2² 1/x1-1/x2 x1-x2用韦达定理怎么变形?
∵x1,x2是方程的·两个根
∴x1x2=c/a=-3/2
x1+x2=-b/a=3
(1)x1x2(x1+x2)
=-3/2*3
=-9/2
(2)1/x1-1/x2=(x2-x1)/x1x2
=√(x2+x1)²-4x1x2/x1x2
=√9+6/(-3/2)
=√15/(-3/2)
=-2√15/2或2√15/2

x1+x2=6/2=3
x1x2=3/2
x1^2x2+x1x2^2
=(x1x2)(x1+x2)
=3*3/2
=9/2
1/x1-1/x2=(x2-x1)/x1x2=2(x2-x1)/3
(x1-x2)^2=x1^2-2x1x2+x2^2
=(x1+x2)^2-4x1x2
=9-4*3/2
=3
1/x1-1/x2=±2√3/3