Question

设x1,x2是方程2x²+4x+-3=0的两个根不解方程求下列各式的值 1.(x1+1)(x2+1) 2. x2／x1+x1／x2

在线等我很着急.

Answer 1

是-3吧？
x1+x2=-2
x1x2=-3/2
所以
(x1+1)(x2+1)
=x1x2+(x1+x2)+1
=-3/2-2+1
=-5/2

x1+x2=-2
两边平方
x1²+2x1x2+x2²=4
x1²+x2²=4-2x1x2=7
所以x2/x1+x1/x2
=(x2²+x1²)/x1x2
=7/(-3/2)
=-14/3

Answer 2

解由题知x1+x2=-2，x1x2=-3/2
故1（x1+1）（x2+1）
=x1x2+（x1+x2）+1
=-3/2-2+1
=-5/2
2 x2/x1+x1/x2
=(x2^2+x1^2)/x1x2
=[(x1+x2)^2-2x1x2]/x1x2
=[(x1+x2)^2]/x1x2-2
=（-2）^2/(-3/2)-2
=-8/3-2
=-14/3.

Answer 3

根据韦达定理有

x1+x2=-b/a=-4/2=-2
x1*x2=c/a=-3/2,则有
1、
(x1+1)(x2+1)
＝x1*x2+x1+x2+1
=-3/2-2+1
=-5/2
2、
x2／x1+x1／x2
=(X2^2+x1^2)/x1*x2
=[(x1+x2)^2-2x1*x2]/x1*x2
=(4+3)/-3/2=-14/3

Answer 4

2x^2+4x-3=0
x1+x2=-2 x1x2=-3/2
(x1+1)(x2+1)=x1x2+x1+x2+1=-3/2-2+1=-5/2
x2/x1+x1/x2=(x1^2+x2^2)/x1x2=[(x1+x2)^2-2x1x2]/x1x2=-14/3