设sin (x+2y-3z)=x+2y-3z，证明sin (x+2y-3z)=x+2y-3z.

sin (x+2y-3z)=x+2y-3z, 则 dfrac (partial z)(partial x)+dfrac (partial z)(partial

5.设 sin (x+2y-3z)=x+2y-3z, 证明: dfrac (partial z)(partial x)+dfrac (partial z)(pa

设平面经过原点及.点 (6,-3,2), 且与平面 4x-y+2z=8 垂直,则此平面方程为 ()-|||-(A) x+2y-3z=0 (B) 2x+y-3z=

设(x,y,z)=(x)^2+(y)^3+z,求(x,y,z)=(x)^2+(y)^3+z,在点(x,y,z)=(x)^2+(y)^3+z,处沿方向(x,y,z

设函数(x,y,z)=2(x)^3y-3(y)^2z在点(x,y,z)=2(x)^3y-3(y)^2z处梯度的模为(x,y,z)=2(x)^3y-3(y)^2z

(x,y,z)=(x)^3+4(y)^3+4(z)^2，则(x,y,z)=(x)^3+4(y)^3+4(z)^2（）。(x,y,z)=(x)^3+4(y)^3+

1.已知 sin (3x-2y+z)=3x-2y+z, 则 dfrac ({partial )^2z}(partial xpartial y)= __

2x+3y+Z=4X-2y+4z=-5(2)3x+8y-2z=134x-y+9z=-62x+3y+Z=4X-2y+4z=-5(2)3x+8y-2z=134x-y

(4) ) 2x+y-z+w=1 3x-2y+z-3w=4 x+4y-3z+5w=-2 .

2x+y-z+W=1(4){3x-2y+z-3w=4x+4y-3z+5w=-22x+y-z+W=1(4){3x-2y+z-3w=4x+4y-3z+5w=-2