设x1,x2是方程2x^2+4x-3=0的两个根,利用根与系数的关系,求下列各式的值.（1）.(x1+3)(x2+3).（2）.x2/x1+x1/x2.

x1，x2是方程2x^2+4x-3=0的两个根
x1+x2=-2
x1*x2=-3/2
(x1+3)(x2+3)
=x1*x2+3(x1+x2)+9
=-3/2+3*(-2)+9
=-3/2-6+9
=3/2
x2/x1+x1/x2
=[(x1)^2+(x2)^2]/(x1*x2)
=[(x1)^2+(x2)^2+2x1*x2-2x1*x2]/(x1*x2)
=[(x1+x2)^2-2x1*x2]/(x1*x2)
=[(-2)^2-2*(-3/2)]/(-3/2)
=(4+3)/(-3/2)
=-7*2/3
=-14/3

x1+x2=-2
x1x2=-3/2
(1)(x1+3)(x2+3)
=x1x2+3(x1+x2)+9
=-3/2-6+9
=3/2
(2)x2/x1+x1/x2
=(x1²+x2²)/x1x2
=[(x1+x2)²-2x1x2]/x1x2
=[4+3]/(-3/2)
=-14/3

x1，x2是方程2x^2+4x-3=0的两个根
则x1+x2=-b/a=-2
x1x2=c/a=-3/2
则
(x1+3)(x2+3)=x1x2+3*(x1+x2)+9=-3/2+3*(-2)+9=3/2
x2/x1+x1/x2=(x2²+x1²)/x1x2=[(x1+x2)²-2x1x2]/x1x2=(x1+x2)²/x1x2-2=4/*(-3/2)-2=-14/3

x1，x2是方程2x^2+4x-3=0的两个根
由韦达定理得：
x1+x2=-4/2=-2
x1x2=-3/2

(1)）(x1+3)(x2+3)
=x1x2+3(x1+x2)+9
=-3/2-6+9
=3/2

（2）x2/x1+x1/x2
=(x1²+x2²)/(x1x2)
=[(x1+x2)²-2x1x2]/(x1x2)
=(x1+x2)²/(x1x2)-2
=4/(-3/2)-2
=-8/3-2
=-14/3

x1,x2是方程2x^2+4x-3=0的两个根
由根与系数的关系得：x1+x2=-4/2=-2,x1*x2=-3/2
（1）(x1+3)(x2+3)=x1*x2+3(x1+x2)+9
=-3/2-6+9
=3/2
（2）通分之后有x1²+x2²出现,先算它
x1²+x2²=(x1+x2)²-2x1*x2
=4+3
=7
x2/x1+x1/x2=(x1²+x2²)/x1*x2=-14/3
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