=\(\frac{4\left(3x-2y\right)}{4^2}\)=\(\frac{3\left(2z-4x\right)}{3^2}\)=\(\frac{2\left(4y-3z\right)}{2^2}\)
= \(\frac{12x-8y}{16}\)=\(\frac{6z-12x}{9}\)=\(\frac{8y-6z}{4}\)
=\(\frac{12x-8y+6z-12x+8y-6z}{16+9+4}\)
=\(\frac{0}{29}\)
=0
+) \(\frac{3x-2y}{4}\)=0 => 3x-2y=0 => 3x=2y => \(\frac{x}{2}\)=\(\frac{y}{3}\) (1)
+) \(\frac{2z-4x}{3}\)=0 => 2z-4x=0 =>2z=4x => \(\frac{z}{4}\)=\(\frac{x}{2}\) (2)
Từ (1) và (2) => \(\frac{x}{2}\)=\(\frac{y}{3}\)=\(\frac{z}{4}\)
tick mình nha