若x1,x2是方程2x^2+5x-1=0的两个根
则:由韦达定理得:
x1+x2=-5/2
x1*x2=-1/2
(1)x1²+x2²=(x1+x2)²-2x1*x2
=(-5/2)²-2×(-1/2)
=25/4+1
=29/4
(2) x2/x1+x1/x2
=(x1²+x2²)/(x1*x2)
=(-5/2)/(-1/2)
=5
若x1,x2是方程2x^2+5x-1=0的两个根
则:由韦达定理得:
x1+x2=-5/2
x1*x2=-1/2
(1)x1²+x2²=(x1+x2)²-2x1*x2
=(-5/2)²-2×(-1/2)
=25/4+1
=29/4
(2) x2/x1+x1/x2
=(x1²+x2²)/(x1*x2)
=(-5/2)/(-1/2)
=5