一道高中数学题，请帮忙解答一下

图

x1+x2= 3
x1.x2=-3/2
(1) (x1)^2.x2 + x1.(x2)^2 = x1.x2.(x1+x2) =-(3/2)(3) =-9/2
(2)
(x1)^2/x2 + (x2)^2/x1
= [(x1)^3+(x2)^3]/ (x1.x2)
=(x1+x2).[( x1+x^2)^2 -3x1.x2 ]/(x1.x2)
=3[ 3^2 +9/2 ] /(-3/2)
=-2[ 9+9/2]
=-27
(3)
(x1-x2)^2
= (x1+x2)^2 -4x1.x2
=9 +6
=15
x1=> x1-x2 = -√15
(x1)^2 -(x2)^2
=(x1 -x2)(x1+x2)
=-3√15
(II)
(x1)^2 -(x2)^2 = -3√15
(x1)^2 < (x2)^2
|x1|<|x2|

-9 3

解1:(1)c=1,2a=4c,a=2c=2

b^2=a^2-c^2=4-1=3

而焦点在x轴上，故椭圆的方程为：x^/4+y^/3=1

(2)设PF1=a,则PF2=4-a

根据余弦定理有：(4-a)^=a^+4-4acos120

即10a=12,PF1=a=6/5

S三角形PF1F2=1/2* PF1*F1F2*sin120=1/2*6/5*2*√3/2=3√3/5