已知2x-3y+z=0,3x-2y-6z=0,xyz不等于0,求x*2+y*2+z*2/2x*2+y*2-z*2
问题描述:
已知2x-3y+z=0,3x-2y-6z=0,xyz不等于0,求x*2+y*2+z*2/2x*2+y*2-z*2
答
由2x-3y+z=0,得:z=3y-2x由3x-2y-6z=0,得z=1/2x-1/3y,于是:3y-2x=1/2x-1/3y,x=3/4y=0.75yz=3/2y=1.5y把x,z的值代入原式中,得[(3/4y)^2+y^2+(3/2y)^2]/[2(3/4y)^2+y^2-(3/2y)^2 ]=[9/16+1+9/4]/[9/8+1-9/4]=(61/16)/...