a²+3a=1b²+3b=1a、b是方程x²+3x-1=0的两根.由韦达定理得a+b=-3 ab=-1b/a²+a/b²=(b³+a³)/(ab)²=[(a+b)(a²-ab+b²)]/(ab)²=(a+b)[(a+b)²-3ab]/(ab)&#...