求直线x-y z 1=0,x y-z-1=0在平面x y z=0上的投影直线方程
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2x-3y-z=0(1)x+3y-14z=0(2)(1)+(2)3x-15z=0x=5z(2)*2-(1)6y-28z+3y+z=09y=27zy=3z代入(4x^2-5xy+z^2)/(xy+yz+
是X+Y/5=Y+X/6=Z+X/7吧由X+Y/5=Y+X/6解得,X=24Y/25把上式代入:Y+X/6=Z+X/7解得Z=179Y/175所以X:Y:Z=(24Y/25):Y:179Y/175=1
题目出错了吧应该是x^2-4x+y^2+6y+√(z-3)+13=0即(x-2)^2+(y+3)^2+√(z-3)=0(xy)^z=(2*(-3))^3=-216啊你也去想想吧
设(y+z)/x=(z+x)/y=(y+x)/z=k则y+z=kx,z+x=ky,y+x=kz三式相加2(x+y+z)=k(x+y+z)故当x+y+z=0时,k=-1,但xy-z不等于0,可知x+y+
Z=f'x(x,y)=xy*[x^(xy-1)]*yZ=f'y(x,y)=xy*[x^(xy-1)]*x再问:答案是Z=f'x(x,y)=yx^xy(lnx+1),Z=f'y(x,y)=x^(xy+1
首先z'(x)=x*(a-x-2*y)=0z'(y)=y(a-y-2*x)=0计算得到四组解(0,0)(a,0)(0,a)(a/3,a/3)1.(0,0)时,f''xx=0,f''xy=a,f''yy
z=xy+x/y对x的偏导数=y+1/y对y的偏导数=x-x/y^2
由2x-3y-z=0,x+3y-14z=0,且x,y,z不全为0解得x=5zy=3z将x=5zy=3z带入4x平方-5XY+Z的平方/xy+yz+zx得4*25z平方-5*5Z*3z+Z的平方/5z*
x^2+y^2+z^2-xy-yz-xz=0(1/2)*2(x^2+y^2+z^2-xy-yz-xz)=0(1/2)*(x^2+y^2-2xy+z^2+y^2-2zy+x^2+z^2-2xz)=0(x
两端对x求偏导得:-ye^(-xy)-2(z/x)+(z/x)e^z=0,所以,z/x=ye^(-xy)/(e^z-2)两端对y求偏导得:-xe^(-xy)-2(z/y)+(z/y)e^z=0,所以,
(x-2)²+(y+3)²+√(z-2)=0则x=2,y=-3,z=2xyz=-12
(x+y+z)²=1²x²+y²+z²+2xy+2yz+2xz=1x²+y²+z²+2(xy+yz+xz)=1x&sup
3x-4y-z=02x+y-8z=08x+4y-32z=03x-4y-z+8x+4y-32z=011x-33z=0x=3zy=2zx²+2xy+z²/xy+yz+zx=(3z)^2
设:改点为(x0,y0)根据距离公式算出平方和,得到一个二元函数表达式,对其求偏导数,得到稳定点,后验证黑塞矩阵的正定或者负定或者不定证明其为极小值点,证毕,得出答案,该点即为所求.(目测是正定从而取
2x-3y-z=0(1)x+3y-14z=0(2)(1)+(2)3x-15z=0x=5z(2)*2-(1)6y-28z+3y+z=09y=27zy=3z代入(4x^2-5xy+z^2)/(xy+yz+
同学,xyz=1吧?这样的话,原式=x/(xy+x+xyz)+y/(yz+y+xyz)+z/(xz+z+xyz)=1/(y+1+yz)+1/(z+1+xz)+1/(x+1+xy)=xyz/(y+xyz
解方程组:{2x-3y-z=0.(1){x+3y-14z=0.(2)(1)+(2)得:3x-15z=0即:x=5z,代入(1)式得y=3z所以:(4x²-5xy+z²)/(xy+y
|x-3|+|y+z|+|2z+1|=0则|x-3|=0x=3|y+z|=0y=-z=1/2|2z+1|=0z=-1/2xy-yz=3x1/2-1/2x(-1/2)=7/4
3x-y=-2zx+2y=-3z那么:x=-z,y=-z(3x^-xy+2y^)/(2x^+4xy+y^)=(3z^2-z^2+2z^2)/(2z^2+4z^2+z^2)=4z^2/7z^2=4/7