Question

设x1,x2是方程3x平方-2x-4=0的两根,不解方程,求下列各式的值，帮我哪。。。。高手帮我。。。

1、x1分之1+x2分之1
2、x1分之x2+x2分之x1
3、（x1-x2）²
要过程。。。。。帮我。。。

Answer 1

由韦达定理 得x1+x2=2/3, x1.x2= -4/3
1、x1分之1+x2分之1=(x1+x2)/x1.x2= -1/2
2、x1分之x2+x2分之x1=(x1+x2)^2/x1.x2-2=-7/3
3、（x1-x2）²=(x1+x2)^2-4x1.x2=52/9

Answer 2

设x1,x2是方程3x平方-2x-4=0的两根,则由根与系数的关系可得：
x1+x2=2/3
x1·x2=-4/3
∴1、x1分之1+x2分之1
=x2/(x1·x2)+x1/(x1·x2)
=(x1+x2)/(x1·x2)
=(2/3)/(-4/3)
=-½
2、x1分之x2+x2分之x1
=[(x2²)/(x1·x2)]+[(x1²)/(x1·x2)]
=(x1²+x2²)/(x1·x2)
=[(x1+x2)²-2x1·x2]/(x1·x2)
=[(2/3)²-2×(-4/3)]/(-4/3)
=(28/9)/(-4/3)
=-7/3
3、（x1-x2）²
=(x1+x2)²-4x1·x2
=(2/3)²-4×(-4/3)
=(4/9)+(16/3)
=52/9

Answer 3

x1+x2=2/3,x1x2=-4/3
1. 1/x1+1/x2=(x1+x2)/x1x2=(2/3)/(-4/3)=-1/2
2. x2/x1+x1/x2=(x1^2+x2^2)/x1x2=((x1+x2)^2-2x1x2)/x1x2=(4/9+8/3)/(-4/3)=-28/12=-7/3
3. (x1-x2)^2=(x1+x2)^2-4x1x2=4/9+16/3=52/9