已知方程组2x-y+z=0,5x+8y-z=0,且xyz≠0,求(2x²-xy+z²)/(2x
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已知方程组2x-y+z=0,5x+8y-z=0,且xyz≠0,求(2x²-xy+z²)/(2x²+y²-z²)的值
2x-y+z=0,(1)
5x+8y-z=0 (2)
(1)*8+(2)
21x+7z=0
21x=-7z
x=-z/3代入(1)
-2z/3-y+z=0
z/3-y=0
y=z/3
所以
(2x²-xy+z²)/(2x²+y²-z²)
=(2z²/9+z²/9+z²)/(2z²/9+z²/9-z²)
=(4z²/3)/(-2z²/3)
=-2
5x+8y-z=0 (2)
(1)*8+(2)
21x+7z=0
21x=-7z
x=-z/3代入(1)
-2z/3-y+z=0
z/3-y=0
y=z/3
所以
(2x²-xy+z²)/(2x²+y²-z²)
=(2z²/9+z²/9+z²)/(2z²/9+z²/9-z²)
=(4z²/3)/(-2z²/3)
=-2
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