求函数f(x,y)=xe-(x²+y²)/2的极值令∂f/∂x=e-x=0,得x=e；∂f/∂y=-y=0,得y=0； 即有驻点(e,0)；A=∂²f/∂x²=-1；B=∂²f/∂x∂y=0；C=ͦ...