Question

Cho \(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}\). Chứng minh:\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)

Answer 1

=\(\frac{4\left(3x-2y\right)}{4^2}\)=\(\frac{3\left(2z-4x\right)}{3^2}\)=\(\frac{2\left(4y-3z\right)}{2^2}\)

= \(\frac{12x-8y}{16}\)=\(\frac{6z-12x}{9}\)=\(\frac{8y-6z}{4}\)

=\(\frac{12x-8y+6z-12x+8y-6z}{16+9+4}\)

=\(\frac{0}{29}\)

=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