椭圆3X^2+2Y^2=12, Z=0. 绕Y轴旋转的旋转面在点M（0,根号3,根号2）处指向外侧的单位法向量为

求f(x,y,z)=2x+y+3z在椭圆x^2+y^2=2,x+z=1的最大最小值 已知点P（x,y）在椭圆X^2/16+y^12=1 .试求z=2x-√3y的最大值 X=3z X+y+z=12 X+2y-2z=10 试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y) 已知(x+y)(x+z)=x,(y+z)(y+x)=2y,(z+x)(z+y)=3z,求x,y,z x=y/z=z/3,x+y+z ＝12,求2x+3y+4z是多少, matlab解对数方程组[x,y,z]=solve('0.29744*x/(3+0.1*x+y+z)=(log10((3+x+y+z)/(3+0.1*x+y+z)))^2','0.79*x/(3+0.1*y+z)=(log10((3+x+y+z)/(3+0.1*x+y+z)*(3+y+z)/(3+0.1*y+z)))^2','1.25*x/(3+0.1*z)=(log10((3+x+y+z)/(3+0.1*x+y+z)*(3+y+z)/(3+0.*1y+z)*(3+z)/ 如果|x+y+z-6|+|2x+3y-z-12|+|2x-y-z|=0求x,y, 3x-y+z=4,2z+3y-z=12,x+y+z=6的解 证明 ：x/(y+z)+y/(z+x)+z/(x+y)>=3/2其中 x,y,z>0 x+2y=3x+2z=4y+z 求x:y:z x/2=y/3=z/5 x+3y-z/x-3y+z x+y/2=z+x/3=y+z/4 x+y+z=27 若x+y+z=3y=2z,则x/x+y+z=? 方程组 (1) x+2y-3z=12 x-3y+2z= - 13 x-y-z=0 (2) x+ y=z 5x-y+z=18 2x+5y+3z= -1 解下列方程组3x-y+z=3,2x+y-3z=11,x+y+z=123x-y+z=32x+y-3z=11x+y+z=12 3x-2y+2z=3,2x+y-3z=11,x+y+z=12 x/2+y+2z=12 x-y/3+z=5 x+y-z/2=3