若x-y+z=0,2x-3y-4z=0,且xyz≠0,求(2x²+3y²+4z²)/(x²+3y²-2z²)的值
问题描述:
若x-y+z=0,2x-3y-4z=0,且xyz≠0,求(2x²+3y²+4z²)/(x²+3y²-2z²)的值
答
由x-y+z=0,2x-3y-4z=0且xyz≠0,
可以得到,
x= -7z,y= -6z
所以
(2x²+3y²+4z²) / (x²+3y²-2z²)
=[2(x/z)² +3(y/z)²+4] / [(x/z)²+3(y/z)² -2] 代入x/z = -7,y/z= -6
=(2×49 +3×36 +4) / (49+3*36 -2)
= 210 / 155
= 42 / 31