填空题 曲面x2+cos(xy)+yz+z=0在点(0,1,-1)处的切平面方程为______.

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【正确答案】 1、x-y+z=-2    
【答案解析】[解析] 令F(x,y,z)=x2+cos(xy)+yz+x,则曲面的法向量
n={F'x,F'y,F'z}={2x-ysin(xy)+1,-xsin(xy)+z,y},
则曲面x2+cos(xy)+yz+x=0在点(0,1,-1)处的法向量为n={1,-1,1},故切平面方程为
(x-0)-(y-1)+(z+1)=0,即x-y+z=-2.