Question

初中代数题

已知x/2=y/3=z/4,求x²-2y²+z²/xy-yz+zx的值

Answer 1

6x=4y=3z
z=2x
y=1.5x
x²-2y²+z²=0.5x²
xy-yz+zx=0.5x²
x²-2y²+z²/xy-yz+zx=1

Answer 2

令a=x/2=y/3=z/4
则x=2a
y=3a
z=4a

所以原式=(4a²-2*9a²+16a²)/(6a²-12a²+8a²)
=2a²/2a²
=1