x1²+3x1+1=0
x1²=-3x1-1
所以x1³=x1(-3x1-1)
=-3x1²-x1
=-3(-3x1-1)-x1
=8x1+3
所以原式=8x1+3+8x2+20
=8(x1+x2)+23
=8×(-3)+23
=-1
kx-2x=1
x=1/(k-2)>0
k-2>0
k>2
x1²+3x1+1=0
x1²=-3x1-1
所以x1³=x1(-3x1-1)
=-3x1²-x1
=-3(-3x1-1)-x1
=8x1+3
所以原式=8x1+3+8x2+20
=8(x1+x2)+23
=8×(-3)+23
=-1
kx-2x=1
x=1/(k-2)>0
k-2>0
k>2