已知4x-3y-6z=0,x+2y-7z=0,xyz≠0,求2x²+3y²+6z²除以x²_数学解答

已知4x-3y-6z=0,x+2y-7z=0,xyz≠0,求2x²+3y²+6z²除以x²+5y²+6z²的值.

人气：264 ℃　时间：2019-08-22 03:31:58

解答

4x-3y-6z=0=>4x-3y=6z (A)
x+2y-7z=0=>x+2y=7z (B)
(B)*3+(A)*2=>11x=33z=>x=3z
(B)*4-(A)=>11y=22z=>y=2z
2x²+3y²+6z²=18z^2+12z^2+6z^2=36z^2
x²+5y²+6z²=9z^2+20z^2+6z^2=35z^2
(2x²+3y²+6z²)/(x²+5y²+6z²)=36z^2/(35z^2)=36/35