试用配方法将二次型f(x,y,z)=x^2+2y^2+5z^2+2xy+2xz+6yz.化为标准形,并写出所有可逆线性变
(2X+Z-Y)/(X^2-XY+XZ-YZ)-(Y-Z)/(X^2-XY-XZ+YZ)
求(2X+Z-Y)/(X^2-XY+XZ-YZ)-(2X+Y+Z)/(X^2+XY+XZ+YZ)
化简(2x-y-z/x^2-xy-xz+yz)+(2y-x-z/y^2-xy-yz+xz)+(2x-x-y/z^2-xz
设x,y,z∈R+,xy+yz+xz=1,证明不等式:(xy)^2/z+(xz)^2/y+(yz)^2/x+6xyz≥x
已知2x-3y-z=0,x+3y-14z=0,且x,y,z不为0.求4x的二次幂-5xy+z的二次幂/xy+yz+xz的
二次型x^2+ay^2+z^2+2bxy+2xz+2yz可经过正交变换(x,y,z)T=P(u,v,w)T化为标准形v^
已知x+y+z=5,xy+yz+xz=7,则x^2+y^2+z^2等于?
x/3=y/2=z/5,求xy+yz+xz/x²+y²+z²
2^x=5^y=10^z证明xy=xz=yz
2^x=5^y=10^z证明xy=xz+yz
解方程组(x+y)/xy=5/6,(y+z)/yz=-2/3,2(x+3)+xz=0
x^2+y^2+xy=25/4,x^2+z^2+xz=169/4,y^2+z^2+yz=36,求xy+xz+yz