1.4x-3y-6z=0,x+2y-7z=0,求(2x^2+3y^2+6z^2)/(x^2+5y^2+7z^2)

Question

1.4x-3y-6z=0,x+2y-7z=0,求(2x^2+3y^2+6z^2)/(x^2+5y^2
+7z^2)

Answer 1

解:由4x-3y-6z=0,(1式)
x+2y-7z=0(2式)
(2式)*4得:
4x+8y-28z=0(3式)
(3式)-(1式)得:
4x+8y-28z-(4x-3y-6z)=0
4x+8y-28z-4x+3y+6z=0
11y-22z=0
y=2z(4式)
将y=2z代入(2式)中得
x+2y-7z=x+2*2z-7z=x-3z=0
x=3z(5式)
将(4式)、(5式)代入
2x^2+3y^2+6z^2/x^2+5y^2+7z^2得：
[2*(3z)^2+3*(2z)^2+6z^2]/[(3z)^2+5*(2z)^2+7z^2]化简得:
=(18z^2+12z^2+6z^2)/(9z^2+20z^2+7z^2)
=36z^2/36z^2
=1